301. For the situation in which a moving mass \(m\) with speed \(u\) sticks to an identical mass at rest, the fraction of initial kinetic energy lost is
ⓐ. \(\frac{1}{4}\)
ⓑ. \(\frac{1}{2}\)
ⓒ. \(\frac{3}{4}\)
ⓓ. \(1\)
Correct Answer: \(\frac{1}{2}\)
Explanation: \( \textbf{Initial kinetic energy:} \)
\[
K_i=\frac{1}{2}mu^2
\]
From momentum conservation for identical masses that stick together,
\[
v=\frac{u}{2}
\]
Final kinetic energy of the combined mass \(2m\) is
\[
K_f=\frac{1}{2}(2m)\left(\frac{u}{2}\right)^2
\]
\[
K_f=m\cdot\frac{u^2}{4}
\]
\[
K_f=\frac{1}{4}mu^2
\]
Compare with
\[
K_i=\frac{1}{2}mu^2
\]
So,
\[
K_f=\frac{1}{2}K_i
\]
The lost fraction is
\[
\frac{K_i-K_f}{K_i}=\frac{1}{2}
\]
Half of the initial kinetic energy is transformed into non-mechanical forms during the sticking collision.
\( \textbf{Final answer:} \) The fraction of initial kinetic energy lost is \(\frac{1}{2}\).
302. Assertion: In a perfectly inelastic collision, the final kinetic energy can be written as \(\frac{p^2}{2M}\), where \(p\) is total momentum and \(M\) is total mass after sticking.
Reason: After sticking, the colliding bodies move together with one common velocity.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: In a perfectly inelastic collision, the bodies stick together and move with a common final velocity. If their total mass is \(M\) and total momentum is \(p\), then \(p=Mv\). Hence, \(v=\frac{p}{M}\). The final kinetic energy is \(K_f=\frac{1}{2}Mv^2\). Substituting \(v=\frac{p}{M}\) gives \(K_f=\frac{p^2}{2M}\). The Reason explains why the final system can be treated as one combined mass.
303. A head-on collision is being checked for consistency. The initial velocities are \(u_1=7\,\text{m s}^{-1}\) and \(u_2=1\,\text{m s}^{-1}\), while the final velocities are \(v_1=2\,\text{m s}^{-1}\) and \(v_2=6\,\text{m s}^{-1}\). The value of \(e\) is
ⓐ. \(\frac{1}{3}\)
ⓑ. \(\frac{2}{3}\)
ⓒ. \(1\)
ⓓ. \(\frac{3}{2}\)
Correct Answer: \(\frac{2}{3}\)
Explanation: \( \textbf{Initial velocities:} \) \(u_1=7\,\text{m s}^{-1}\) and \(u_2=1\,\text{m s}^{-1}\).
\( \textbf{Final velocities:} \) \(v_1=2\,\text{m s}^{-1}\) and \(v_2=6\,\text{m s}^{-1}\).
Relative speed of approach is
\[
u_1-u_2=7-1=6\,\text{m s}^{-1}
\]
Relative speed of separation is
\[
v_2-v_1=6-2=4\,\text{m s}^{-1}
\]
Coefficient of restitution is
\[
e=\frac{v_2-v_1}{u_1-u_2}
\]
\[
e=\frac{4}{6}=\frac{2}{3}
\]
The value is less than \(1\), so this is not a perfectly elastic collision.
\( \textbf{Final answer:} \) The value of \(e\) is \(\frac{2}{3}\).
304. A table gives total kinetic energy before and after collisions of isolated two-body systems.
| Case | \(K_{\text{before}}\) | \(K_{\text{after}}\) | Best classification |
|---|
| P | \(40\,\text{J}\) | \(40\,\text{J}\) | elastic |
| Q | \(40\,\text{J}\) | \(30\,\text{J}\) | inelastic |
| R | \(40\,\text{J}\) | \(0\,\text{J}\) | possible if total momentum is zero and bodies stick |
| S | \(40\,\text{J}\) | \(30\,\text{J}\) | elastic |
The row that needs correction is
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row S
Explanation: In an isolated collision, momentum is conserved, but kinetic energy decides whether the collision is elastic or inelastic. Row P is correct because equal total kinetic energies before and after indicate an elastic collision. Row Q is correct because a decrease in kinetic energy indicates an inelastic collision. Row R can occur in a perfectly inelastic collision if the total momentum is zero, so the stuck mass remains at rest after collision. Row S is incorrect because a drop from \(40\,\text{J}\) to \(30\,\text{J}\) cannot be classified as elastic.
305. A smooth ball hits a fixed wall obliquely. If the wall is vertical and smooth, and the collision is elastic, the component of velocity parallel to the wall
ⓐ. reverses direction
ⓑ. remains unchanged
ⓒ. becomes zero
ⓓ. becomes twice its initial value
Correct Answer: remains unchanged
Explanation: For a smooth wall, the impulsive force during collision acts normal to the wall. Therefore, the velocity component perpendicular to the wall is affected by the collision. The component parallel to the wall has no impulsive force acting along it, so it remains unchanged. In an elastic collision with a fixed smooth wall, the normal component reverses sign with the same magnitude. This component-wise view is useful because velocity is a vector, while kinetic energy depends on the combined speed.
306. A ball approaches a smooth vertical wall with velocity components \(v_x=6\,\text{m s}^{-1}\) perpendicular to the wall and \(v_y=8\,\text{m s}^{-1}\) parallel to the wall. If the collision is elastic, the speed after rebound is
ⓐ. \(6\,\text{m s}^{-1}\)
ⓑ. \(8\,\text{m s}^{-1}\)
ⓒ. \(10\,\text{m s}^{-1}\)
ⓓ. \(14\,\text{m s}^{-1}\)
Correct Answer: \(10\,\text{m s}^{-1}\)
Explanation: \( \textbf{Perpendicular component before collision:} \) \(v_x=6\,\text{m s}^{-1}\).
\( \textbf{Parallel component before collision:} \) \(v_y=8\,\text{m s}^{-1}\).
For an elastic collision with a fixed smooth wall, the perpendicular component reverses direction but keeps its magnitude.
So after collision,
\[
v_x'=-6\,\text{m s}^{-1}
\]
The parallel component remains unchanged:
\[
v_y'=8\,\text{m s}^{-1}
\]
The speed after rebound is
\[
v'=\sqrt{(v_x')^2+(v_y')^2}
\]
\[
v'=\sqrt{(-6)^2+8^2}
\]
\[
v'=\sqrt{36+64}
\]
\[
v'=10\,\text{m s}^{-1}
\]
The direction changes, but the speed remains the same in this ideal elastic wall collision.
\( \textbf{Final answer:} \) The speed after rebound is \(10\,\text{m s}^{-1}\).
307. The following statements refer to oblique collision with a fixed smooth wall.
I. The velocity component normal to the wall is affected by the collision.
II. The velocity component parallel to the wall remains unchanged for a smooth wall.
III. In an elastic collision, the speed remains unchanged.
IV. In an elastic collision, both components must become zero.
The true statements are
ⓐ. I, II, and III only
ⓑ. I and IV only
ⓒ. II, III, and IV only
ⓓ. I, II, III, and IV
Correct Answer: I, II, and III only
Explanation: The impulsive force from a smooth wall acts normal to the wall, so the normal component of velocity changes. There is no tangential impulsive force for a smooth wall, so the parallel component remains unchanged. In an elastic collision with a fixed smooth wall, the normal component reverses with the same magnitude, so the total speed remains unchanged. Statement IV is false because the ball rebounds rather than losing both velocity components. The wall changes the direction of motion, not the mass or the entire velocity magnitude in the ideal elastic case.
308. A collision is analysed using momentum conservation, kinetic energy comparison, and coefficient of restitution. Which summary is most accurate?
ⓐ. Use momentum alone to identify whether the collision is elastic
ⓑ. kinetic energy or \(e\) to classify, and momentum for the system constraint
ⓒ. Use coefficient of restitution instead of momentum in all collisions
ⓓ. Use short contact time as proof that kinetic energy is conserved
Correct Answer: kinetic energy or \(e\) to classify, and momentum for the system constraint
Explanation: For an isolated collision, total momentum conservation gives a required relation between velocities before and after collision. Kinetic energy comparison tells whether the collision is elastic or inelastic. The coefficient of restitution describes the ratio of relative separation speed to relative approach speed and also helps classify the collision. None of these ideas should be confused with the others: momentum conservation is about external impulse, kinetic energy conservation is about energy loss or recovery, and \(e\) describes rebound behavior. A short interaction time alone does not guarantee kinetic energy conservation.
309. A collision between two isolated bodies has total kinetic energy \(60\,\text{J}\) before collision and \(45\,\text{J}\) after collision. The energy change during the collision is best described as
ⓐ. \(15\,\text{J}\) becomes non-mechanical energy
ⓑ. \(15\,\text{J}\) of momentum is lost from the system
ⓒ. \(45\,\text{J}\) of kinetic energy is newly created
ⓓ. \(60\,\text{J}\) of kinetic energy remains unchanged
Correct Answer: \(15\,\text{J}\) becomes non-mechanical energy
Explanation: In an isolated collision, total momentum is conserved, but total kinetic energy need not be conserved. Here the kinetic energy decreases from \(60\,\text{J}\) to \(45\,\text{J}\). The loss in kinetic energy is \(60\,\text{J}-45\,\text{J}=15\,\text{J}\). This lost kinetic energy is transformed into internal energy, deformation, heat, or sound. The decrease in kinetic energy does not mean that total momentum is lost.
310. A collision table is given below.
| Case | Total kinetic energy before | Total kinetic energy after | Energy classification |
|---|
| P | \(80\,\text{J}\) | \(80\,\text{J}\) | elastic |
| Q | \(80\,\text{J}\) | \(50\,\text{J}\) | inelastic with kinetic-energy loss |
| R | \(80\,\text{J}\) | \(95\,\text{J}\) | kinetic-energy gain from internal energy |
| S | \(80\,\text{J}\) | \(50\,\text{J}\) | perfectly elastic |
The row that needs correction is
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row S
Explanation: Row P is correct because equal kinetic energy before and after is the elastic case. Row Q is correct because a decrease in kinetic energy indicates an inelastic collision. Row R can describe a separation or explosive type of event where internal energy is converted into kinetic energy. Row S is incorrect because kinetic energy decreases from \(80\,\text{J}\) to \(50\,\text{J}\), so the collision cannot be perfectly elastic. Elastic classification requires total kinetic energy to remain unchanged.
311. A two-body isolated collision has initial kinetic energy \(120\,\text{J}\). After collision, the total kinetic energy is \(90\,\text{J}\). The value of \(K_{\text{after}}-K_{\text{before}}\) is
ⓐ. \(+30\,\text{J}\)
ⓑ. \(-30\,\text{J}\)
ⓒ. \(+210\,\text{J}\)
ⓓ. \(-210\,\text{J}\)
Correct Answer: \(-30\,\text{J}\)
Explanation: \( \textbf{Initial kinetic energy:} \)
\[
K_{\text{before}}=120\,\text{J}
\]
\( \textbf{Final kinetic energy:} \)
\[
K_{\text{after}}=90\,\text{J}
\]
The change in kinetic energy is
\[
\Delta K=K_{\text{after}}-K_{\text{before}}
\]
Substitute:
\[
\Delta K=90-120
\]
\[
\Delta K=-30\,\text{J}
\]
The negative sign means the system has less kinetic energy after collision.
The missing \(30\,\text{J}\) has changed into non-mechanical forms such as internal energy, heat, deformation, or sound.
\( \textbf{Final answer:} \) \(K_{\text{after}}-K_{\text{before}}=-30\,\text{J}\).
312. A collision-like event has no external impulse, and the total kinetic energy after the event is greater than before. The most suitable interpretation is that
ⓐ. momentum conservation must fail
ⓑ. internal energy has changed into kinetic energy
ⓒ. the event must be a perfectly elastic collision
ⓓ. kinetic energy cannot increase in an isolated system
Correct Answer: internal energy has changed into kinetic energy
Explanation: If external impulse is negligible, total momentum of the system can remain conserved. Kinetic energy, however, can increase if some stored internal energy is converted into kinetic energy. An explosion or spring-release separation can behave this way. This does not make the event elastic, because elastic means kinetic energy remains the same, not that it increases. The increase in kinetic energy comes from another energy store within the system.
313. Read the situation below and answer the question.
Two small carts move on a frictionless horizontal track. During a short interaction, a compressed spring between them is released. No external horizontal impulse acts on the two-cart system during the release.
The total momentum of the two carts and their total kinetic energy after release are best described as
ⓐ. momentum conserved, kinetic energy may increase
ⓑ. momentum not conserved, kinetic energy must remain constant
ⓒ. momentum conserved, kinetic energy must be zero
ⓓ. momentum not conserved, kinetic energy must decrease
Correct Answer: momentum conserved, kinetic energy may increase
Explanation: The absence of external horizontal impulse means the total horizontal momentum of the two-cart system is conserved. The spring is internal to the system and can convert stored elastic potential energy into kinetic energy. Therefore, the total kinetic energy of the carts may increase after the spring is released. This does not violate energy conservation, because the added kinetic energy comes from spring potential energy. Momentum conservation and kinetic-energy conservation are separate conditions.
314. In an isolated event, two objects initially at rest push apart because of internal energy release. Their momenta just after separation must be
ⓐ. equal and opposite in direction
ⓑ. equal in magnitude and same in direction
ⓒ. zero for both parts after separation
ⓓ. unrelated because kinetic energy increases
Correct Answer: equal and opposite in direction
Explanation: Initially, the two-object system is at rest, so its total momentum is zero. If no external impulse acts, total momentum remains zero after separation. Therefore, the two final momenta must add to zero. This requires equal magnitudes and opposite directions. The kinetic energy can increase because internal energy is transformed into kinetic energy, but that does not change the zero total momentum condition.
315. A body initially at rest explodes into two fragments of masses \(2\,\text{kg}\) and \(3\,\text{kg}\). The \(2\,\text{kg}\) fragment moves to the right with speed \(6\,\text{m s}^{-1}\). The velocity of the \(3\,\text{kg}\) fragment is
ⓐ. \(+4\,\text{m s}^{-1}\)
ⓑ. \(-4\,\text{m s}^{-1}\)
ⓒ. \(+9\,\text{m s}^{-1}\)
ⓓ. \(-9\,\text{m s}^{-1}\)
Correct Answer: \(-4\,\text{m s}^{-1}\)
Explanation: \( \textbf{Initial condition:} \) The body is initially at rest, so total initial momentum is zero.
Take right as positive.
For the \(2\,\text{kg}\) fragment:
\[
p_1=(2)(+6)=+12\,\text{kg m s}^{-1}
\]
Let the velocity of the \(3\,\text{kg}\) fragment be \(v_2\).
Momentum conservation gives
\[
p_1+p_2=0
\]
\[
12+3v_2=0
\]
\[
3v_2=-12
\]
\[
v_2=-4\,\text{m s}^{-1}
\]
The negative sign means the \(3\,\text{kg}\) fragment moves to the left.
\( \textbf{Final answer:} \) The velocity is \(-4\,\text{m s}^{-1}\).
316. A stationary object separates into two parts. Part P has mass \(m\), and part Q has mass \(3m\). If part P moves with speed \(6v\), the speed of part Q is
ⓐ. \(2v\)
ⓑ. \(3v\)
ⓒ. \(6v\)
ⓓ. \(18v\)
Correct Answer: \(2v\)
Explanation: The initial momentum of the object is zero because it is stationary. After separation, total momentum must still be zero if no external impulse acts. Therefore, the magnitudes of the two momenta must be equal:
\[
m(6v)=(3m)V
\]
Cancel \(m\):
\[
6v=3V
\]
Solve for \(V\):
\[
V=2v
\]
The heavier part moves more slowly so that its momentum can balance the lighter part's momentum.
\( \textbf{Final answer:} \) The speed of part Q is \(2v\).
317. Two fragments separate from rest with equal and opposite momenta. Fragment P has mass \(1\,\text{kg}\), and fragment Q has mass \(4\,\text{kg}\). The ratio of their kinetic energies \(K_P:K_Q\) is
ⓐ. \(1:4\)
ⓑ. \(4:1\)
ⓒ. \(1:1\)
ⓓ. \(2:1\)
Correct Answer: \(4:1\)
Explanation: Since the fragments separate from rest, their momenta are equal in magnitude and opposite in direction.
For the same momentum magnitude, kinetic energy is
\[
K=\frac{p^2}{2m}
\]
Thus, kinetic energy is inversely proportional to mass when \(p\) is fixed.
For P and Q:
\[
\frac{K_P}{K_Q}=\frac{p^2/(2m_P)}{p^2/(2m_Q)}
\]
\[
\frac{K_P}{K_Q}=\frac{m_Q}{m_P}
\]
Substitute \(m_P=1\,\text{kg}\) and \(m_Q=4\,\text{kg}\):
\[
\frac{K_P}{K_Q}=\frac{4}{1}
\]
The lighter fragment gets more kinetic energy in the separation.
\( \textbf{Final answer:} \) \(K_P:K_Q=4:1\).
318. An object at rest explodes into two fragments. One fragment has kinetic energy \(36\,\text{J}\), while the other has mass twice as large. If no external impulse acts, the kinetic energy of the heavier fragment is
ⓐ. \(18\,\text{J}\)
ⓑ. \(36\,\text{J}\)
ⓒ. \(72\,\text{J}\)
ⓓ. \(144\,\text{J}\)
Correct Answer: \(18\,\text{J}\)
Explanation: Since the object was initially at rest, the two fragments must have equal and opposite momenta after explosion. For a fixed momentum magnitude,
\[
K=\frac{p^2}{2m}
\]
So kinetic energy is inversely proportional to mass.
Let the lighter fragment have mass \(m\) and kinetic energy \(36\,\text{J}\).
The heavier fragment has mass \(2m\).
Therefore, its kinetic energy is half as large:
\[
K_{\text{heavy}}=\frac{36}{2}
\]
\[
K_{\text{heavy}}=18\,\text{J}
\]
The heavier piece moves more slowly and carries less kinetic energy for the same momentum magnitude.
\( \textbf{Final answer:} \) The heavier fragment has kinetic energy \(18\,\text{J}\).
319. Use the before-after information below.
| Stage | Information |
|---|
| Before separation | Two-part system at rest, total momentum \(0\) |
| After separation | Part P moves right; part Q moves left |
| Energy source | Internal stored energy of the system |
The best conclusion is
ⓐ. momentum remains zero while kinetic energy is produced
ⓑ. total momentum becomes positive because one part moves right
ⓒ. kinetic energy stays zero because the system began at rest
ⓓ. the heavier part must always have greater kinetic energy
Correct Answer: momentum remains zero while kinetic energy is produced
Explanation: The initial total momentum is zero. If the separation occurs without external impulse, momentum conservation requires the final total momentum to remain zero. This is possible when the two parts have equal and opposite momenta. Kinetic energy can be non-zero after separation because internal stored energy has been converted into motion. The heavier part does not necessarily have greater kinetic energy; for equal momentum magnitudes, the lighter part has greater kinetic energy.
320. A \(4\,\text{kg}\) object at rest separates into two fragments of masses \(1\,\text{kg}\) and \(3\,\text{kg}\). If the \(1\,\text{kg}\) fragment has speed \(9\,\text{m s}^{-1}\), the total kinetic energy after separation is
ⓐ. \(27\,\text{J}\)
ⓑ. \(54\,\text{J}\)
ⓒ. \(72\,\text{J}\)
ⓓ. \(108\,\text{J}\)
Correct Answer: \(54\,\text{J}\)
Explanation: \( \textbf{Initial momentum:} \) The object is initially at rest, so total momentum is zero.
Take the \(1\,\text{kg}\) fragment's direction as positive.
Momentum of the \(1\,\text{kg}\) fragment is
\[
p_1=(1)(9)=9\,\text{kg m s}^{-1}
\]
The \(3\,\text{kg}\) fragment must have momentum
\[
p_2=-9\,\text{kg m s}^{-1}
\]
So its velocity is
\[
v_2=\frac{-9}{3}=-3\,\text{m s}^{-1}
\]
Kinetic energy of the first fragment:
\[
K_1=\frac{1}{2}(1)(9)^2=40.5\,\text{J}
\]
Kinetic energy of the second fragment:
\[
K_2=\frac{1}{2}(3)(3)^2=13.5\,\text{J}
\]
Total kinetic energy is
\[
K_{\text{total}}=40.5+13.5=54\,\text{J}
\]
This kinetic energy comes from internal energy released during separation.
\( \textbf{Final answer:} \) The total kinetic energy after separation is \(54\,\text{J}\).
