501. Consider a one-dimensional perfectly inelastic collision between two bodies of positive masses.
Statement I: The common final velocity is a mass-weighted average of the two initial signed velocities.
Statement II: If both initial velocities are equal, sticking them together causes no kinetic-energy loss associated with relative collision motion.
Statement III: For a projectile of mass \(m\) striking a stationary target of mass \(M\), the fraction of initial kinetic energy retained after sticking is \(m/(m+M)\).
Statement IV: For a fixed projectile mass and speed, increasing the stationary target mass increases the common final speed.
Which of the following is correct?
ⓐ. Statements I and III only
ⓑ. Statements I, II and III only
ⓒ. Statements II, III and IV only
ⓓ. Statements I, II, III and IV
Correct Answer: Statements I, II and III only
Explanation: Statement I follows directly from momentum conservation:
\[
V=\frac{m_1u_1+m_2u_2}{m_1+m_2}.
\]
Because the masses are positive, this is a mass-weighted average of the signed velocities.
Statement II is correct.
If
\[
u_1=u_2=u,
\]
then the common final velocity is already \(u\), so there is no change in translational kinetic energy due to sticking.
Statement III is also correct for a stationary target:
\[
V=\frac{m}{m+M}u.
\]
This gives
\[
\frac{K_f}{K_i}=\frac{m}{m+M}.
\]
Statement IV is incorrect.
Increasing \(M\) enlarges the denominator in
\[
V=\frac{m}{m+M}u,
\]
so the common final speed decreases.
Hence Statements I, II and III are correct.
502.
A \(1\,\text{kg}\) body moving horizontally at \(10\,\text{m s}^{-1}\) strikes and sticks to a stationary \(1\,\text{kg}\) compact body located at the lowest point of a vertical circle. The target is attached to a light string of length \(0.40\,\text{m}\). The impact is along the tangent at the lowest point, and external impulse along this tangent during the short collision is negligible. After collision the stuck pair moves as a compact particle on the string. Take \(g=10\,\text{m s}^{-2}\).
Which set gives the speed immediately after collision, the kinetic energy lost in the collision, and the tension at the highest point if the pair completes the circle?
ⓐ. \(4\,\text{m s}^{-1},\;34\,\text{J},\;20\,\text{N}\)
ⓑ. \(5\,\text{m s}^{-1},\;25\,\text{J},\;0\)
ⓒ. \(5\,\text{m s}^{-1},\;25\,\text{J},\;25\,\text{N}\)
ⓓ. \(5\,\text{m s}^{-1},\;20\,\text{J},\;45\,\text{N}\)
Correct Answer: \(5\,\text{m s}^{-1},\;25\,\text{J},\;25\,\text{N}\)
Explanation: \(\textbf{Tangential momentum during the sticking collision:}\)
\[
(1)(10)=(1+1)V
\]
Hence
\[
V=5\,\text{m s}^{-1}
\]
The initial kinetic energy is
\[
K_i=\frac12(1)(10^2)=50\,\text{J}
\]
Immediately after sticking,
\[
K_f=\frac12(2)(5^2)=25\,\text{J}
\]
Therefore the collision loses
\[
K_{\text{lost}}=25\,\text{J}
\]
\(\textbf{Check the vertical-circle constraint:}\)
At the bottom,
\[
V^2=25\,\text{m}^2\text{s}^{-2}
\]
The minimum value for complete taut-string motion is
\[
5gR=5(10)(0.40)=20\,\text{m}^2\text{s}^{-2}
\]
Since
\[
25\gt 20,
\]
the pair can complete the circle.
\(\textbf{Top speed from mechanical energy:}\)
\[
v_t^2=V^2-4gR
\]
Thus
\[
v_t^2=25-4(10)(0.40)=9
\]
At the top, with total mass \(2\,\text{kg}\),
\[
T_t+Mg=\frac{Mv_t^2}{R}
\]
Therefore
\[
T_t+20=\frac{(2)(9)}{0.40}
\]
Hence
\[
T_t+20=45,
\]
so
\[
T_t=25\,\text{N}
\]
\(\textbf{Final answer:}\)
\[
\boxed{V=5\,\text{m s}^{-1},\quad K_{\text{lost}}=25\,\text{J},\quad T_t=25\,\text{N}}
\]
503. Two identical bodies of mass \(m\) undergo an isolated one-dimensional collision. Initially body 1 has momentum \(p\) and body 2 is stationary. After collision their momenta in the original direction are \(p/4\) and \(3p/4\), respectively. Which statement correctly classifies the collision and gives the fraction of initial kinetic energy lost?
ⓐ. Elastic; no kinetic energy is lost
ⓑ. Inelastic with separation; \(3/8\) of the initial kinetic energy is lost
ⓒ. Perfectly inelastic; \(1/2\) of the initial kinetic energy is lost
ⓓ. Inelastic with separation; \(5/8\) of the initial kinetic energy is lost
Correct Answer: Inelastic with separation; \(3/8\) of the initial kinetic energy is lost
Explanation: The total final momentum is
\[
\frac p4+\frac{3p}{4}=p,
\]
so momentum conservation is satisfied.
For a body of mass \(m\), kinetic energy in terms of momentum is
\[
K=\frac{p^2}{2m}.
\]
Thus the initial kinetic energy is
\[
K_i=\frac{p^2}{2m}
\]
\(\textbf{Final system kinetic energy:}\)
\[
K_f
=
\frac{(p/4)^2}{2m}
+
\frac{(3p/4)^2}{2m}
\]
Therefore
\[
K_f
=
\frac{p^2}{2m}
\left(
\frac1{16}+\frac9{16}
\right)
\]
Hence
\[
K_f=\frac58K_i
\]
The fractional kinetic-energy loss is therefore
\[
1-\frac58=\frac38
\]
The final momenta are unequal, so the bodies are not moving together.
Moreover, the initially stationary front body acquires the greater forward speed, allowing separation after collision.
Thus the outcome is inelastic but not perfectly inelastic.
\(\textbf{Final answer:}\)
\[
\text{inelastic with separation},\qquad \frac{K_{\text{lost}}}{K_i}=\frac38.
\]
504. Consider a collision in a plane for which the external impulse during the interaction is negligible.
Statement I: Conservation of momentum is a vector condition, so the total momentum must be conserved independently along two perpendicular axes.
Statement II: For an elastic collision, conservation of total kinetic energy supplies an additional scalar condition besides the momentum-component equations.
Statement III: It is sufficient to conserve only the magnitude of the total momentum; its direction may change without any external impulse.
Statement IV: In a general two-dimensional collision, momentum conservation alone may not be sufficient to determine all unknown final speeds and directions.
Which of the following is correct?
ⓐ. Statements I and III only
ⓑ. Statements I, II and IV only
ⓒ. Statements II, III and IV only
ⓓ. Statements I, II, III and IV
Correct Answer: Statements I, II and IV only
Explanation: Statement I is correct because momentum is a vector quantity.
For an isolated collision,
\[
\vec p_i=\vec p_f.
\]
Therefore one may write separate conservation equations along \(x\) and \(y\).
Statement II is also correct.
For an elastic collision the additional requirement is
\[
K_i=K_f.
\]
This energy equation is scalar and is independent of the two momentum-component equations.
Statement III is incorrect because equal momentum magnitudes with different directions correspond to different momentum vectors.
Without external impulse, the total momentum vector itself cannot change.
Statement IV is correct because several final speeds and angles may remain unknown even after writing the two component equations.
Hence additional information such as elasticity, a final speed or a final angle may be required.
505.
A \(1\,\text{kg}\) particle moves initially at \(5\,\text{m s}^{-1}\) along the \(+x\)-axis and collides with an identical stationary particle. After collision, particle 1 moves at \(3\,\text{m s}^{-1}\) above the \(+x\)-axis, with \(\cos\theta=3/5\) and \(\sin\theta=4/5\). External impulse is negligible.
Which set correctly gives the speed and direction of particle 2 and classifies the collision?
ⓐ. \(4\,\text{m s}^{-1}\), \(36.9^\circ\) below \(+x\); elastic
ⓑ. \(4\,\text{m s}^{-1}\), \(36.9^\circ\) above \(+x\); inelastic
ⓒ. \(3\,\text{m s}^{-1}\), \(53.1^\circ\) below \(+x\); elastic
ⓓ. \(5\,\text{m s}^{-1}\), along \(+x\); perfectly inelastic
Correct Answer: \(4\,\text{m s}^{-1}\), \(36.9^\circ\) below \(+x\); elastic
Explanation: \(\textbf{Resolve the measured velocity of particle 1:}\)
\[
v_{1x}=3\left(\frac35\right)=\frac95\,\text{m s}^{-1},
\]
\[
v_{1y}=3\left(\frac45\right)=\frac{12}{5}\,\text{m s}^{-1}
\]
The initial momentum is
\[
\vec p_i=5\hat i
\]
\(\textbf{Use momentum conservation in the }x\textbf{-direction:}\)
\[
5=\frac95+v_{2x}
\]
Hence
\[
v_{2x}=\frac{16}{5}\,\text{m s}^{-1}
\]
\(\textbf{Use momentum conservation in the }y\textbf{-direction:}\)
\[
0=\frac{12}{5}+v_{2y}
\]
Thus
\[
v_{2y}=-\frac{12}{5}\,\text{m s}^{-1}
\]
Therefore
\[
v_2=\sqrt{\left(\frac{16}{5}\right)^2+\left(\frac{12}{5}\right)^2}=4\,\text{m s}^{-1}
\]
Its direction satisfies
\[
\tan\phi=\frac{12}{16}=\frac34,
\]
so
\[
\phi\approx36.9^\circ
\]
below the \(+x\)-axis.
\(\textbf{Energy classification:}\)
Initially,
\[
K_i=\frac12(1)(5^2)=12.5\,\text{J}
\]
Finally,
\[
K_f=\frac12(1)(3^2)+\frac12(1)(4^2)=4.5+8=12.5\,\text{J}
\]
Both vector momentum and total kinetic energy are conserved.
\(\textbf{Final answer:}\)
\[
v_2=4\,\text{m s}^{-1}\text{ at }36.9^\circ\text{ below }+x;\ \text{elastic}.
\]
506. Assertion: In an elastic two-dimensional collision between two equal masses, if one mass is initially at rest and both final speeds are non-zero, the two final velocity vectors are perpendicular.
Reason: Momentum conservation gives
\[
\vec u=\vec v_1+\vec v_2,
\]
while kinetic-energy conservation gives
\[
u^2=v_1^2+v_2^2;
\]
squaring the momentum equation therefore requires \(\vec v_1\cdot\vec v_2=0\).
ⓐ. Both Assertion and Reason are correct, and the Reason correctly explains the Assertion
ⓑ. Both Assertion and Reason are correct, but the Reason does not correctly explain the Assertion
ⓒ. Assertion is correct, but Reason is incorrect
ⓓ. Assertion is incorrect, but Reason is correct
Correct Answer: Both Assertion and Reason are correct, and the Reason correctly explains the Assertion
Explanation: For equal masses with the target initially at rest, momentum conservation becomes
\[
\vec u=\vec v_1+\vec v_2.
\]
Taking the square of this vector relation,
\[
u^2=v_1^2+v_2^2+2\vec v_1\cdot\vec v_2.
\]
Because the collision is elastic,
\[
u^2=v_1^2+v_2^2.
\]
Comparing the two equations gives
\[
2\vec v_1\cdot\vec v_2=0.
\]
Hence
\[
\vec v_1\cdot\vec v_2=0.
\]
For two non-zero final velocities this means that the angle between them is \(90^\circ\).
The conclusion depends on the equal-mass, elastic and initially stationary-target conditions.
The Reason therefore correctly explains the Assertion.
507.
A particle of mass \(m\) moving at \(10\,\text{m s}^{-1}\) along the \(+x\)-axis collides elastically with an identical stationary particle. After collision, the incident particle moves at \(30^\circ\) above the \(+x\)-axis. Both final speeds are non-zero.
What are the two final speeds and the direction of the second particle?
ⓐ. \(10\,\text{m s}^{-1}\) and \(0\); second particle remains at rest
ⓑ. \(5\,\text{m s}^{-1}\) and \(5\sqrt3\,\text{m s}^{-1}\); second particle at \(30^\circ\) below \(+x\)
ⓒ. \(5\sqrt2\,\text{m s}^{-1}\) each; second particle at \(45^\circ\) below \(+x\)
ⓓ. \(5\sqrt3\,\text{m s}^{-1}\) and \(5\,\text{m s}^{-1}\); second particle at \(60^\circ\) below \(+x\)
Correct Answer: \(5\sqrt3\,\text{m s}^{-1}\) and \(5\,\text{m s}^{-1}\); second particle at \(60^\circ\) below \(+x\)
Explanation: For equal masses with one initially at rest in an elastic collision, the non-zero final velocity vectors are perpendicular.
Therefore, if particle 1 emerges at
\[
30^\circ
\]
above \(+x\), particle 2 must move at
\[
60^\circ
\]
below \(+x\).
Let their speeds be \(v_1\) and \(v_2\).
\(\textbf{Transverse momentum balance:}\)
\[
v_1\sin30^\circ=v_2\sin60^\circ
\]
Thus
\[
\frac{v_1}{2}=\frac{\sqrt3}{2}v_2,
\]
so
\[
v_1=\sqrt3\,v_2
\]
\(\textbf{Longitudinal momentum balance:}\)
\[
10=v_1\cos30^\circ+v_2\cos60^\circ
\]
Substituting,
\[
10=(\sqrt3v_2)\frac{\sqrt3}{2}+v_2\frac12
\]
Hence
\[
10=2v_2
\]
Therefore
\[
v_2=5\,\text{m s}^{-1},
\]
and
\[
v_1=5\sqrt3\,\text{m s}^{-1}
\]
A check gives
\[
v_1^2+v_2^2=75+25=100=u^2,
\]
as required for an elastic equal-mass collision.
\(\textbf{Final answer:}\)
\[
\boxed{v_1=5\sqrt3\,\text{m s}^{-1},\quad v_2=5\,\text{m s}^{-1},\quad \phi_2=60^\circ\text{ below }+x}
\]
508.
A \(2\,\text{kg}\) particle moves initially at \(3\,\text{m s}^{-1}\) along the \(+x\)-axis and collides with a stationary \(1\,\text{kg}\) particle. After collision the \(2\,\text{kg}\) particle has velocity \((2\hat i+\hat j)\,\text{m s}^{-1}\). External impulse is negligible.
Which statement is correct?
ⓐ. The \(1\,\text{kg}\) particle has velocity \((2\hat i-2\hat j)\,\text{m s}^{-1}\); the collision is elastic, but the two final velocity vectors are not perpendicular
ⓑ. The \(1\,\text{kg}\) particle has velocity \((2\hat i-2\hat j)\,\text{m s}^{-1}\); the collision is inelastic because the final velocities are not perpendicular
ⓒ. The \(1\,\text{kg}\) particle has velocity \((4\hat i-2\hat j)\,\text{m s}^{-1}\); the collision is elastic
ⓓ. The \(1\,\text{kg}\) particle has velocity \((2\hat i+2\hat j)\,\text{m s}^{-1}\); the collision is perfectly inelastic
Correct Answer: The \(1\,\text{kg}\) particle has velocity \((2\hat i-2\hat j)\,\text{m s}^{-1}\); the collision is elastic, but the two final velocity vectors are not perpendicular
Explanation: \(\textbf{Initial momentum vector:}\)
\[
\vec p_i=(2)(3)\hat i=6\hat i
\]
The final momentum of the \(2\,\text{kg}\) particle is
\[
\vec p_1=2(2\hat i+\hat j)=4\hat i+2\hat j
\]
Therefore momentum conservation requires
\[
\vec p_2=6\hat i-(4\hat i+2\hat j)
\]
Hence
\[
\vec p_2=2\hat i-2\hat j
\]
Since the second mass is \(1\,\text{kg}\),
\[
\vec v_2=(2\hat i-2\hat j)\,\text{m s}^{-1}
\]
\(\textbf{Check total kinetic energy:}\)
Initially,
\[
K_i=\frac12(2)(3^2)=9\,\text{J}
\]
Finally, for particle 1,
\[
K_1=\frac12(2)(2^2+1^2)=5\,\text{J}
\]
For particle 2,
\[
K_2=\frac12(1)(2^2+(-2)^2)=4\,\text{J}
\]
Thus
\[
K_f=5+4=9\,\text{J}
\]
The collision is elastic.
However,
\[
(2,1)\cdot(2,-2)=4-2=2\neq0
\]
Thus the final velocity vectors are not perpendicular.
The familiar \(90^\circ\) result does not apply because the masses are unequal.
\(\textbf{Final answer:}\)
\[
\vec v_2=(2\hat i-2\hat j)\,\text{m s}^{-1};\ \text{elastic but not perpendicular}.
\]
509. Two identical \(1\,\text{kg}\) particles collide in a plane. Initially particle 1 moves at \(5\,\text{m s}^{-1}\) along \(+x\), while particle 2 is at rest. The measured final velocities are
\[
\vec v_1=(2\hat i+\hat j)\,\text{m s}^{-1},
\qquad
\vec v_2=(3\hat i-\hat j)\,\text{m s}^{-1}.
\]
Which set correctly classifies the collision and gives the kinetic-energy loss?
ⓐ. Elastic; \(0\,\text{J}\)
ⓑ. Impossible because transverse momentum is not conserved
ⓒ. Perfectly inelastic; \(5\,\text{J}\)
ⓓ. Inelastic; \(5\,\text{J}\)
Correct Answer: Inelastic; \(5\,\text{J}\)
Explanation: \(\textbf{Check vector momentum first:}\)
Initially,
\[
\vec p_i=5\hat i
\]
Finally,
\[
\vec p_f=(2\hat i+\hat j)+(3\hat i-\hat j)
\]
Hence
\[
\vec p_f=5\hat i
\]
Thus both the \(x\)- and \(y\)-momentum conditions are satisfied.
\(\textbf{Compare total kinetic energies:}\)
Initially,
\[
K_i=\frac12(1)(5^2)=12.5\,\text{J}
\]
Finally,
\[
K_f=\frac12(2^2+1^2)+\frac12(3^2+1^2)
\]
Therefore
\[
K_f=\frac52+5=7.5\,\text{J}
\]
The kinetic-energy decrease is
\[
K_{\text{lost}}=12.5-7.5=5\,\text{J}
\]
The final velocities are different, so the particles have not stuck together.
Hence the collision is inelastic but not perfectly inelastic.
Also,
\[
\vec v_1\cdot\vec v_2=6-1=5\neq0,
\]
showing that the equal-mass right-angle result cannot be used for this inelastic collision.
\(\textbf{Final answer:}\)
\[
\text{inelastic collision with }5\,\text{J kinetic-energy loss}.
\]
510.
A \(1\,\text{kg}\) projectile moves initially along the \(+x\)-axis and strikes a stationary target of unknown mass \(M\). After collision the projectile velocity is \((2\hat i+2\hat j)\,\text{m s}^{-1}\), while the target velocity is \((\hat i-\hat j)\,\text{m s}^{-1}\). External impulse is negligible.
What are \(M\), the projectile's initial speed, and the kinetic-energy loss?
ⓐ. \(M=2\,\text{kg},\;u=6\,\text{m s}^{-1},\;4\,\text{J}\)
ⓑ. \(M=1\,\text{kg},\;u=3\,\text{m s}^{-1},\;0\)
ⓒ. \(M=2\,\text{kg},\;u=4\,\text{m s}^{-1},\;2\,\text{J}\)
ⓓ. \(M=4\,\text{kg},\;u=6\,\text{m s}^{-1},\;2\,\text{J}\)
Correct Answer: \(M=2\,\text{kg},\;u=4\,\text{m s}^{-1},\;2\,\text{J}\)
Explanation: The initial transverse momentum is zero.
\(\textbf{Use the }y\textbf{-momentum equation to infer the target mass:}\)
\[
0=(1)(2)+M(-1)
\]
Therefore
\[
M=2\,\text{kg}
\]
\(\textbf{Use the }x\textbf{-momentum equation to infer the incident speed:}\)
\[
(1)u=(1)(2)+(2)(1)
\]
Hence
\[
u=4\,\text{m s}^{-1}
\]
\(\textbf{Initial kinetic energy:}\)
\[
K_i=\frac12(1)(4^2)=8\,\text{J}
\]
\(\textbf{Final kinetic energy of the projectile:}\)
\[
K_1=\frac12(1)(2^2+2^2)=4\,\text{J}
\]
For the target,
\[
K_2=\frac12(2)(1^2+(-1)^2)=2\,\text{J}
\]
Thus
\[
K_f=6\,\text{J}
\]
Therefore
\[
K_{\text{lost}}=8-6=2\,\text{J}
\]
Momentum is conserved vectorially, but total kinetic energy decreases, so the collision is inelastic.
\(\textbf{Final answer:}\)
\[
\boxed{M=2\,\text{kg},\qquad u=4\,\text{m s}^{-1},\qquad K_{\text{lost}}=2\,\text{J}}
\]
511.
A \(2\,\text{kg}\) particle moves with velocity \(3\hat i\,\text{m s}^{-1}\), while a \(1\,\text{kg}\) particle moves with velocity \(4\hat j\,\text{m s}^{-1}\). They collide and stick together. External impulse during collision is negligible.
Which set gives the speed of the composite body, the tangent of the angle its velocity makes with \(+x\), and the kinetic energy lost?
ⓐ. \(\dfrac{2\sqrt{13}}{3}\,\text{m s}^{-1},\;\dfrac23,\;\dfrac{25}{3}\,\text{J}\)
ⓑ. \(\sqrt{13}\,\text{m s}^{-1},\;\dfrac32,\;\dfrac{13}{3}\,\text{J}\)
ⓒ. \(\dfrac{2\sqrt{13}}{3}\,\text{m s}^{-1},\;\dfrac32,\;\dfrac{25}{6}\,\text{J}\)
ⓓ. \(2\,\text{m s}^{-1},\;\dfrac23,\;5\,\text{J}\)
Correct Answer: \(\dfrac{2\sqrt{13}}{3}\,\text{m s}^{-1},\;\dfrac23,\;\dfrac{25}{3}\,\text{J}\)
Explanation: \(\textbf{Vector momentum before sticking:}\)
\[
\vec p_i=(2)(3)\hat i+(1)(4)\hat j
\]
Thus
\[
\vec p_i=6\hat i+4\hat j
\]
The combined mass after collision is
\[
M=3\,\text{kg}
\]
Therefore the common velocity is
\[
\vec V=\frac{\vec p_i}{3}
\]
Hence
\[
\vec V=2\hat i+\frac43\hat j
\]
Its speed is
\[
V=\sqrt{2^2+\left(\frac43\right)^2}
\]
Thus
\[
V=\frac{2\sqrt{13}}{3}\,\text{m s}^{-1}
\]
The direction satisfies
\[
\tan\theta=\frac{4/3}{2}=\frac23
\]
\(\textbf{Energy lost in the sticking collision:}\)
Initially,
\[
K_i=\frac12(2)(3^2)+\frac12(1)(4^2)
\]
\[
K_i=9+8=17\,\text{J}
\]
Finally,
\[
K_f=\frac12(3)\left(\frac{52}{9}\right)=\frac{26}{3}\,\text{J}
\]
Therefore
\[
K_{\text{lost}}=17-\frac{26}{3}=\frac{25}{3}\,\text{J}
\]
\(\textbf{Final answer:}\)
\[
\boxed{V=\frac{2\sqrt{13}}{3}\,\text{m s}^{-1},\quad \tan\theta=\frac23,\quad K_{\text{lost}}=\frac{25}{3}\,\text{J}}
\]
512. Two identical \(1\,\text{kg}\) particles undergo separate collision trials. In every trial, particle 1 initially moves at \(6\,\text{m s}^{-1}\) along \(+x\) and particle 2 is stationary.
| Record | \(\vec v_1\) after collision | \(\vec v_2\) after collision |
|---|
| P | \((2\hat i+2\hat j)\,\text{m s}^{-1}\) | \((4\hat i-2\hat j)\,\text{m s}^{-1}\) |
| Q | \((3\hat i+3\hat j)\,\text{m s}^{-1}\) | \((3\hat i-3\hat j)\,\text{m s}^{-1}\) |
| R | \((5\hat i+3\hat j)\,\text{m s}^{-1}\) | \((\hat i-3\hat j)\,\text{m s}^{-1}\) |
Which classification is correct for an isolated collision with no additional stored energy source?
ⓐ. P elastic, Q inelastic, R impossible
ⓑ. P inelastic, Q elastic, R impossible
ⓒ. P impossible, Q elastic, R inelastic
ⓓ. P inelastic, Q impossible, R elastic
Correct Answer: P inelastic, Q elastic, R impossible
Explanation: The initial momentum is
\[
\vec p_i=6\hat i\,\text{kg m s}^{-1},
\]
and the initial kinetic energy is
\[
K_i=\frac12(6^2)=18\,\text{J}.
\]
All three records satisfy
\[
\vec v_1+\vec v_2=6\hat i,
\]
so vector momentum alone does not distinguish them.
\(\textbf{Record P:}\)
\[
K_P=\frac12(2^2+2^2)+\frac12(4^2+(-2)^2).
\]
Thus
\[
K_P=4+10=14\,\text{J}.
\]
Momentum is conserved but kinetic energy decreases, so P is inelastic.
\(\textbf{Record Q:}\)
\[
K_Q=\frac12(3^2+3^2)+\frac12(3^2+(-3)^2).
\]
Therefore
\[
K_Q=9+9=18\,\text{J}.
\]
Hence Q is elastic.
\(\textbf{Record R:}\)
\[
K_R=\frac12(5^2+3^2)+\frac12(1^2+(-3)^2).
\]
Thus
\[
K_R=17+5=22\,\text{J}.
\]
This exceeds the initial \(18\,\text{J}\).
For the stated isolated collision with no additional energy source, record R is impossible.
\(\textbf{Final answer:}\)
\[
\text{P inelastic,\ Q elastic,\ R impossible}.
\]
513. A target is initially at rest and a projectile of known mass and known speed moves along the \(+x\)-axis. After a general two-dimensional elastic collision, suppose both final speeds and both final directions are unknown. Which statement about the available equations is correct?
ⓐ. The two momentum-component equations alone determine all four unknowns
ⓑ. Momentum in \(x\) and \(y\) plus kinetic-energy conservation give three independent equations; one more independent datum is generally required
ⓒ. Kinetic-energy conservation alone determines both final speeds and both angles
ⓓ. Momentum in \(x\) and \(y\) plus kinetic-energy conservation give four independent equations, so no further datum is required
Correct Answer: Momentum in \(x\) and \(y\) plus kinetic-energy conservation give three independent equations; one more independent datum is generally required
Explanation: Let the unknown final quantities be
\[
v_1,\quad v_2,\quad\theta_1,\quad\theta_2.
\]
There are therefore four scalar unknowns.
Vector momentum conservation provides two scalar equations:
\[
p_{ix}=p_{fx},
\]
and
\[
p_{iy}=p_{fy}.
\]
Because the collision is elastic, total kinetic-energy conservation supplies one additional scalar equation:
\[
K_i=K_f.
\]
Thus conservation laws provide three independent scalar relations for four unknowns.
The equations therefore generally leave one degree of freedom.
An additional independent measurement or condition, such as one final angle or one final speed, is ordinarily required.
In the special equal-mass stationary-target case one can derive perpendicular final velocities, but that relation itself follows from the same momentum-and-energy equations and does not create an extra independent conservation law.
514. In an elastic collision between two identical masses, one mass is initially at rest and the other moves along \(+x\). After collision the faster body has speed \(2v\) and moves above the \(+x\)-axis, while the slower body has speed \(v\) and moves below it. Both final speeds are non-zero. Which set is correct?
ⓐ. Faster-body angle \(=\tan^{-1}(1/2)\), slower-body angle \(=\tan^{-1}(2)\), incident speed \(=\sqrt5\,v\)
ⓑ. Faster-body angle \(=45^\circ\), slower-body angle \(=45^\circ\), incident speed \(=3v\)
ⓒ. Faster-body angle \(=\tan^{-1}(2)\), slower-body angle \(=\tan^{-1}(1/2)\), incident speed \(=\sqrt3\,v\)
ⓓ. Faster-body angle \(=30^\circ\), slower-body angle \(=60^\circ\), incident speed \(=2v\)
Correct Answer: Faster-body angle \(=\tan^{-1}(1/2)\), slower-body angle \(=\tan^{-1}(2)\), incident speed \(=\sqrt5\,v\)
Explanation: For equal masses with one initially stationary in an elastic collision, the two final velocity vectors are perpendicular.
Let the faster-body angle above \(+x\) be \(\alpha\), and the slower-body angle below \(+x\) be \(\beta\).
Therefore
\[
\alpha+\beta=90^\circ
\]
\(\textbf{Transverse momentum cancellation:}\)
\[
(2v)\sin\alpha=v\sin\beta
\]
Using
\[
\sin\beta=\cos\alpha,
\]
we obtain
\[
2\sin\alpha=\cos\alpha
\]
Hence
\[
\tan\alpha=\frac12
\]
Therefore
\[
\alpha=\tan^{-1}\left(\frac12\right),
\]
and
\[
\beta=\tan^{-1}(2)
\]
\(\textbf{Incident speed from energy conservation:}\)
\[
u^2=(2v)^2+v^2
\]
Thus
\[
u^2=5v^2,
\]
so
\[
u=\sqrt5\,v
\]
\(\textbf{Final answer:}\)
\[
\boxed{\alpha=\tan^{-1}(1/2),\quad\beta=\tan^{-1}(2),\quad u=\sqrt5\,v}
\]
515.
A \(1\,\text{kg}\) block is released from an ideal spring of force constant \(400\,\text{N m}^{-1}\), initially compressed by \(0.50\,\text{m}\), on a smooth horizontal surface. At the spring's natural length it collides elastically with an identical stationary block. After collision the original block moves at \(45^\circ\) above the original line of motion. The second block then moves onto a smooth incline. Take \(g=10\,\text{m s}^{-2}\).
What are the speed of the second block immediately after collision, its direction, and the maximum vertical height it reaches?
ⓐ. \(10\,\text{m s}^{-1}\), along the original direction, \(5\,\text{m}\)
ⓑ. \(5\,\text{m s}^{-1}\), \(45^\circ\) below the original direction, \(1.25\,\text{m}\)
ⓒ. \(5\sqrt2\,\text{m s}^{-1}\), \(45^\circ\) below the original direction, \(2.5\,\text{m}\)
ⓓ. \(5\sqrt2\,\text{m s}^{-1}\), \(90^\circ\) to the original direction, \(1.25\,\text{m}\)
Correct Answer: \(5\sqrt2\,\text{m s}^{-1}\), \(45^\circ\) below the original direction, \(2.5\,\text{m}\)
Explanation: \(\textbf{Spring-to-collision speed:}\)
The initial spring energy is
\[
U_s=\frac12kx^2
\]
Substituting,
\[
U_s=\frac12(400)(0.50)^2=50\,\text{J}
\]
On the smooth surface this becomes kinetic energy:
\[
\frac12(1)u^2=50
\]
Hence
\[
u=10\,\text{m s}^{-1}
\]
\(\textbf{Two-dimensional elastic collision:}\)
The masses are equal and the target is initially at rest.
Thus the two non-zero final velocity vectors are perpendicular.
The original block moves at \(45^\circ\) above the initial direction, so the target must move at \(45^\circ\) below it.
Because their transverse momenta cancel and the angles have equal magnitudes, the two final speeds are equal.
Let each speed be \(v\).
Energy conservation gives
\[
10^2=v^2+v^2
\]
Hence
\[
v=5\sqrt2\,\text{m s}^{-1}
\]
\(\textbf{Climb of the second block:}\)
Its kinetic energy converts into gravitational potential energy:
\[
\frac12v^2=gh
\]
Thus
\[
\frac12(50)=10h
\]
Therefore
\[
h=2.5\,\text{m}
\]
\(\textbf{Final answer:}\)
\[
\boxed{v=5\sqrt2\,\text{m s}^{-1},\quad45^\circ\text{ below the original direction},\quad h_{\max}=2.5\,\text{m}}
\]
516.
A \(1\,\text{kg}\) block is released from rest at an unknown height \(h\) on a smooth track. At the bottom it collides elastically in a plane with an identical stationary block. After collision the two blocks move on opposite sides of the original direction with speeds \(8\,\text{m s}^{-1}\) and \(6\,\text{m s}^{-1}\), respectively. Take \(g=10\,\text{m s}^{-2}\).
If the \(8\,\text{m s}^{-1}\) block moves above the original direction, which set gives \(h\) and the two scattering angles?
ⓐ. \(h=5\,\text{m};\quad36.9^\circ\) above and \(53.1^\circ\) below
ⓑ. \(h=4\,\text{m};\quad53.1^\circ\) above and \(36.9^\circ\) below
ⓒ. \(h=5\,\text{m};\quad45^\circ\) above and \(45^\circ\) below
ⓓ. \(h=2.5\,\text{m};\quad36.9^\circ\) above and \(53.1^\circ\) below
Correct Answer: \(h=5\,\text{m};\quad36.9^\circ\) above and \(53.1^\circ\) below
Explanation: For equal masses with one initially at rest in an elastic collision, the two final velocity vectors are perpendicular.
Therefore the incident speed \(u\) satisfies
\[
u^2=8^2+6^2
\]
Hence
\[
u^2=100,
\]
so
\[
u=10\,\text{m s}^{-1}
\]
\(\textbf{Recover the release height:}\)
The pre-collision speed resulted from gravitational descent:
\[
mgh=\frac12mu^2
\]
Thus
\[
10h=\frac12(100)
\]
Hence
\[
h=5\,\text{m}
\]
\(\textbf{Determine the directions from transverse momentum:}\)
Let the \(8\,\text{m s}^{-1}\) block move at angle \(\alpha\) above \(+x\), while the \(6\,\text{m s}^{-1}\) block moves at \(\beta\) below it.
The perpendicular geometry gives
\[
\alpha+\beta=90^\circ
\]
Transverse momentum conservation gives
\[
8\sin\alpha=6\sin\beta
\]
Using
\[
\sin\beta=\cos\alpha,
\]
\[
8\sin\alpha=6\cos\alpha
\]
Therefore
\[
\tan\alpha=\frac34
\]
Hence
\[
\alpha\approx36.9^\circ,
\qquad
\beta\approx53.1^\circ
\]
\(\textbf{Final answer:}\)
\[
\boxed{h=5\,\text{m},\quad\alpha=36.9^\circ,\quad\beta=53.1^\circ}
\]
517. Two identical \(1\,\text{kg}\) particles are involved in a collision experiment. Before collision particle 1 moves at \(6\,\text{m s}^{-1}\) along \(+x\), while particle 2 is at rest. Three different measured final records are shown.
| Trial | \(\vec v_1\) | \(\vec v_2\) |
|---|
| P | \((4\hat i+\hat j)\,\text{m s}^{-1}\) | \((2\hat i-\hat j)\,\text{m s}^{-1}\) |
| Q | \((3\hat i+3\hat j)\,\text{m s}^{-1}\) | \((3\hat i-3\hat j)\,\text{m s}^{-1}\) |
| R | \((5\hat i+3\hat j)\,\text{m s}^{-1}\) | \((\hat i-3\hat j)\,\text{m s}^{-1}\) |
Which statement is correct?
ⓐ. P and Q are elastic; R is inelastic
ⓑ. P is inelastic, Q is elastic, and R requires an additional energy source
ⓒ. P is perfectly inelastic, Q is elastic, and R is inelastic
ⓓ. All three are allowed elastic outcomes because the momentum vectors add correctly
Correct Answer: P is inelastic, Q is elastic, and R requires an additional energy source
Explanation: The initial momentum is
\[
\vec p_i=6\hat i,
\]
and the initial kinetic energy is
\[
K_i=18\,\text{J}.
\]
Each record has zero net \(y\)-momentum and total \(x\)-momentum \(6\,\text{kg m s}^{-1}\).
Thus every row passes the momentum test.
\(\textbf{Trial P energy:}\)
\[
K_P=\frac12(4^2+1^2)+\frac12(2^2+1^2).
\]
Therefore
\[
K_P=\frac{17}{2}+\frac52=11\,\text{J}.
\]
Hence P is inelastic.
\(\textbf{Trial Q energy:}\)
\[
K_Q=\frac12(18)+\frac12(18)=18\,\text{J}.
\]
Thus Q is elastic.
\(\textbf{Trial R energy:}\)
\[
K_R=\frac12(34)+\frac12(10)=22\,\text{J}.
\]
This exceeds the initial \(18\,\text{J}\).
Therefore R cannot arise from the stated isolated collision unless some additional internal or external energy source, not given in the problem, supplies energy.
The record demonstrates why vector momentum conservation alone is not enough to classify a collision.
\(\textbf{Final answer:}\)
\[
\text{P inelastic,\ Q elastic,\ R inconsistent with the stated isolated collision}.
\]
518.
Two \(1\,\text{kg}\) particles move on a smooth horizontal plane. Before collision, particle A has velocity \(4\hat i\,\text{m s}^{-1}\) and particle B has velocity \(3\hat j\,\text{m s}^{-1}\). They collide and stick together. The composite body then immediately engages an ideal horizontal spring of force constant \(200\,\text{N m}^{-1}\) on a smooth surface.
Which set gives the direction of the composite velocity, the kinetic energy lost in the collision, and the maximum spring compression?
ⓐ. \(\tan\theta=3/4,\;\dfrac{25}{4}\,\text{J},\;0.25\,\text{m}\)
ⓑ. \(\tan\theta=4/3,\;\dfrac{25}{4}\,\text{J},\;0.50\,\text{m}\)
ⓒ. \(\tan\theta=3/4,\;5\,\text{J},\;0.25\,\text{m}\)
ⓓ. \(\tan\theta=1,\;\dfrac{25}{2}\,\text{J},\;0.50\,\text{m}\)
Correct Answer: \(\tan\theta=3/4,\;\dfrac{25}{4}\,\text{J},\;0.25\,\text{m}\)
Explanation: \(\textbf{Vector momentum of the sticking pair:}\)
Initially,
\[
\vec p_i=4\hat i+3\hat j
\]
The total mass after collision is
\[
2\,\text{kg}
\]
Therefore
\[
\vec V=\frac{4\hat i+3\hat j}{2}
\]
Hence
\[
\vec V=2\hat i+\frac32\hat j
\]
The direction satisfies
\[
\tan\theta=\frac{3/2}{2}=\frac34
\]
The speed is
\[
V=\sqrt{2^2+\left(\frac32\right)^2}=\frac52\,\text{m s}^{-1}
\]
\(\textbf{Collision energy loss:}\)
Initially,
\[
K_i=\frac12(1)(4^2)+\frac12(1)(3^2)
\]
Thus
\[
K_i=8+\frac92=\frac{25}{2}\,\text{J}
\]
Immediately after sticking,
\[
K_f=\frac12(2)\left(\frac52\right)^2
\]
Therefore
\[
K_f=\frac{25}{4}\,\text{J}
\]
The kinetic energy lost during collision is
\[
K_{\text{lost}}=\frac{25}{2}-\frac{25}{4}=\frac{25}{4}\,\text{J}
\]
\(\textbf{Spring-energy stage:}\)
The remaining translational kinetic energy is stored in the spring:
\[
\frac12(200)x^2=\frac{25}{4}
\]
Thus
\[
100x^2=\frac{25}{4}
\]
Hence
\[
x^2=\frac1{16},
\]
so
\[
x=0.25\,\text{m}
\]
\(\textbf{Final answer:}\)
\[
\boxed{\tan\theta=\frac34,\qquad K_{\text{lost}}=\frac{25}{4}\,\text{J},\qquad x_{\max}=0.25\,\text{m}}
\]
519. Consider a collision in a plane for which the net external impulse during the short interaction is negligible.
Statement I: Conservation of momentum requires conservation of the total \(x\)-momentum and total \(y\)-momentum separately.
Statement II: In an elastic collision, conservation of total kinetic energy provides an additional scalar condition beyond the two momentum-component equations.
Statement III: If the magnitudes of total momentum before and after collision are equal, momentum conservation is automatically satisfied even if their directions differ.
Statement IV: In a perfectly inelastic collision, the velocity of the composite body is directed along the total initial momentum vector.
Which of the following is correct?
ⓐ. Statements I and III only
ⓑ. Statements I, II and IV only
ⓒ. Statements II, III and IV only
ⓓ. Statements I, II, III and IV
Correct Answer: Statements I, II and IV only
Explanation: Momentum is a vector quantity.
For negligible external impulse,
\[
\vec p_i=\vec p_f.
\]
Hence the component equations
\[
p_{ix}=p_{fx}
\]
and
\[
p_{iy}=p_{fy}
\]
must both be satisfied, so Statement I is correct.
An elastic collision must additionally satisfy
\[
K_i=K_f,
\]
making Statement II correct.
Statement III is incorrect because two vectors can have equal magnitudes but different directions.
Such vectors are not equal and therefore do not represent conserved momentum.
For a perfectly inelastic collision,
\[
(m_1+m_2)\vec V=\vec p_i.
\]
Thus \(\vec V\) is parallel to the total initial momentum, so Statement IV is correct.
520. A particle of mass \(m\) moving along the \(+x\)-axis collides elastically with an identical stationary particle. Both particles move after collision. The incident particle retains \(64\%\) of its initial kinetic energy. Which set correctly gives the fraction of initial kinetic energy acquired by the target and the directions of the two final velocities?
ⓐ. \(36\%\); incident particle at \(36.9^\circ\) and target at \(53.1^\circ\) on opposite sides of \(+x\)
ⓑ. \(36\%\); incident particle at \(53.1^\circ\) and target at \(36.9^\circ\) on opposite sides of \(+x\)
ⓒ. \(64\%\); both particles at \(45^\circ\) on opposite sides of \(+x\)
ⓓ. \(20\%\); incident particle at \(36.9^\circ\) and target at \(53.1^\circ\)
Correct Answer: \(36\%\); incident particle at \(36.9^\circ\) and target at \(53.1^\circ\) on opposite sides of \(+x\)
Explanation: Let the incident speed before collision be \(u\).
Since the incident particle retains \(64\%\) of its kinetic energy,
\[
\frac{v_1^2}{u^2}=0.64
\]
Hence
\[
v_1=0.8u
\]
Because the collision is elastic, the target receives the remaining \(36\%\):
\[
\frac{v_2^2}{u^2}=0.36,
\]
so
\[
v_2=0.6u
\]
For equal masses with one initially at rest, the two non-zero final velocity vectors are perpendicular.
Thus the momentum vectors form a right triangle whose hypotenuse has magnitude \(u\).
If \(\alpha\) is the deflection angle of the incident particle,
\[
\cos\alpha=\frac{v_1}{u}=0.8
\]
Therefore
\[
\alpha\approx36.9^\circ
\]
The target direction is perpendicular to it and lies on the opposite side of the original direction:
\[
\beta=90^\circ-\alpha\approx53.1^\circ
\]
\(\textbf{Final answer:}\)
\[
36\%;\quad36.9^\circ\text{ and }53.1^\circ
\]
