601.
A particle of mass \(m\) is released from rest at height \(H\) on a smooth track and collides head-on elastically with an identical stationary particle at the lowest point of a vertical circle of radius \(R\). The struck particle moves on a light string around the circle. The string can sustain a maximum tension of \(8mg\).
For what range of \(H\) can the struck particle complete the circle without the string becoming slack or breaking? Also, what is the top tension at the upper end of this range?
ⓐ. \(\dfrac{5R}{2}\leq H\leq\dfrac{7R}{2}\), and \(T_t=2mg\)
ⓑ. \(2R\leq H\leq\dfrac{5R}{2}\), and \(T_t=mg\)
ⓒ. \(\dfrac{5R}{2}\leq H\leq4R\), and \(T_t=3mg\)
ⓓ. \(5R\leq H\leq7R\), and \(T_t=2mg\)
Correct Answer: \(\dfrac{5R}{2}\leq H\leq\dfrac{7R}{2}\), and \(T_t=2mg\)
Explanation: In the equal-mass elastic collision, the projectile stops and the initially stationary target acquires the projectile speed \(u\).
The smooth descent gives
\[
u^2=2gH
\]
\(\textbf{Lower bound from the taut-string completion condition:}\)
For a complete vertical circle,
\[
u^2\geq5gR
\]
Therefore
\[
2gH\geq5gR
\]
Hence
\[
H\geq\frac{5R}{2}
\]
\(\textbf{Upper bound from the breaking strength:}\)
The maximum string tension occurs at the bottom:
\[
T_b=mg+\frac{mu^2}{R}
\]
The strength condition is
\[
T_b\leq8mg
\]
Thus
\[
mg+\frac{mu^2}{R}\leq8mg,
\]
giving
\[
u^2\leq7gR
\]
Therefore
\[
H\leq\frac{7R}{2}
\]
At the upper limit,
\[
u^2=7gR
\]
The top speed satisfies
\[
v_t^2=u^2-4gR=3gR
\]
At the top,
\[
T_t+mg=\frac{m(3gR)}{R}
\]
Hence
\[
T_t=2mg
\]
\(\textbf{Final answer:}\)
\[
\boxed{\frac{5R}{2}\leq H\leq\frac{7R}{2},\qquad T_t=2mg\text{ at }H=\frac{7R}{2}}
\]
602.
A \(1\,\text{kg}\) projectile of unknown speed \(u\) collides head-on with a stationary \(2\,\text{kg}\) target. After collision the projectile rebounds and is stopped by a constant \(2\,\text{N}\) resistive force over \(1\,\text{m}\). The target moves forward and is stopped by a constant \(4\,\text{N}\) resistive force over \(4\,\text{m}\). No other force does work during the stopping stages.
Which set correctly gives the incident speed, the magnitude of impulse on the projectile, and the collision type?
ⓐ. \(8\,\text{m s}^{-1},\;10\,\text{N s},\) inelastic
ⓑ. \(6\,\text{m s}^{-1},\;4\,\text{N s},\) inelastic
ⓒ. \(4\,\text{m s}^{-1},\;6\,\text{N s},\) elastic
ⓓ. \(6\,\text{m s}^{-1},\;8\,\text{N s},\) elastic
Correct Answer: \(6\,\text{m s}^{-1},\;8\,\text{N s},\) elastic
Explanation: \(\textbf{Recover the projectile's post-collision speed from stopping work:}\)
The resistive force removes
\[
W_1=(2)(1)=2\,\text{J}
\]
Hence
\[
\frac12(1)v_1^2=2
\]
Thus
\[
|v_1|=2\,\text{m s}^{-1}
\]
The projectile rebounds, so
\[
v_1=-2\,\text{m s}^{-1}
\]
\(\textbf{Recover the target speed:}\)
The target loses
\[
W_2=(4)(4)=16\,\text{J}
\]
Therefore
\[
\frac12(2)v_2^2=16
\]
Hence
\[
v_2=4\,\text{m s}^{-1}
\]
\(\textbf{Infer the incident speed from momentum:}\)
\[
(1)u=(1)(-2)+(2)(4)
\]
Thus
\[
u=6\,\text{m s}^{-1}
\]
The projectile impulse is
\[
J=(1)|-2-6|=8\,\text{N s}
\]
\(\textbf{Classify the collision:}\)
Initially,
\[
K_i=\frac12(1)(6^2)=18\,\text{J}
\]
Finally,
\[
K_f=2+16=18\,\text{J}
\]
Hence total kinetic energy is conserved.
\(\textbf{Final answer:}\)
\[
u=6\,\text{m s}^{-1},\qquad |J|=8\,\text{N s},\qquad \text{elastic collision}.
\]
603. A projectile of mass \(m\) and speed \(u\) strikes a stationary target of mass \(3m\). Trial E is head-on and perfectly elastic. Trial P is perfectly inelastic. In Trial P, after the collision is over, a motor accelerates the stuck \(4m\) body along the same direction until its speed becomes equal to the speed acquired by the \(3m\) target in Trial E. Which statement is correct?
ⓐ. The motor work equals the kinetic energy lost in the perfectly inelastic collision
ⓑ. The motor work is half the kinetic energy lost in the perfectly inelastic collision
ⓒ. The motor work is twice the kinetic energy lost in the perfectly inelastic collision
ⓓ. The motor does zero net work because momentum was conserved during collision
Correct Answer: The motor work equals the kinetic energy lost in the perfectly inelastic collision
Explanation: \(\textbf{Target speed in Trial E:}\)
For the elastic collision,
\[
v_E=\frac{2m}{m+3m}u=\frac u2.
\]
\(\textbf{Common speed immediately after Trial P:}\)
For the perfectly inelastic collision,
\[
mu=(4m)V.
\]
Hence
\[
V=\frac u4.
\]
The kinetic energy immediately after sticking is
\[
K_P=\frac12(4m)\left(\frac u4\right)^2.
\]
Thus
\[
K_P=\frac18mu^2.
\]
The initial kinetic energy was
\[
K_i=\frac12mu^2.
\]
Therefore the collision loss is
\[
K_{\text{lost}}
=
\frac12mu^2-\frac18mu^2
=
\frac38mu^2.
\]
\(\textbf{Motor work after the collision:}\)
The motor raises the speed of the \(4m\) composite from \(u/4\) to \(u/2\).
Hence
\[
W_M
=
\frac12(4m)\left(\frac u2\right)^2
-
\frac12(4m)\left(\frac u4\right)^2.
\]
Therefore
\[
W_M
=
\frac12mu^2-\frac18mu^2
=
\frac38mu^2.
\]
Thus
\[
W_M=K_{\text{lost}}.
\]
The later motor work does not retroactively make the earlier collision elastic.
\(\textbf{Final answer:}\)
\[
\boxed{W_M=K_{\text{lost}}=\frac38mu^2}
\]
604. Two identical particles undergo an isolated two-dimensional elastic collision. Both particles are moving before collision. After collision, particle 1 is observed to be at rest while particle 2 is moving. Which additional condition must have been satisfied by the two initial velocity vectors?
ⓐ. They were parallel and in the same direction
ⓑ. They were parallel and opposite in direction
ⓒ. They were perpendicular
ⓓ. They had equal magnitudes, irrespective of direction
Correct Answer: They were perpendicular
Explanation: Let the two initial velocity vectors be
\[
\vec u_1,\qquad\vec u_2.
\]
Because the masses are identical, momentum conservation gives
\[
\vec u_1+\vec u_2=\vec v_2
\]
when particle 1 finally comes to rest.
Elasticity requires
\[
u_1^2+u_2^2=v_2^2.
\]
Using the momentum relation,
\[
v_2^2=|\vec u_1+\vec u_2|^2.
\]
Therefore
\[
u_1^2+u_2^2
=
u_1^2+u_2^2+2\vec u_1\cdot\vec u_2.
\]
Hence
\[
\vec u_1\cdot\vec u_2=0.
\]
Thus the two initial velocity vectors must have been perpendicular.
This is a useful converse to the familiar equal-mass stationary-target right-angle result.
605.
Two identical \(1\,\text{kg}\) particles have initial velocities \(6\hat i\,\text{m s}^{-1}\) and \(8\hat j\,\text{m s}^{-1}\). The same initial state is used in two trials. In Trial E the collision is elastic and particle 1 is at rest immediately afterward. In Trial P the collision is perfectly inelastic. In each trial, the moving post-collision body subsequently compresses an ideal spring of force constant \(200\,\text{N m}^{-1}\) on a smooth surface.
Which set gives the maximum spring compressions in Trial E and Trial P?
ⓐ. \(\dfrac{1}{\sqrt2}\,\text{m},\;\dfrac{1}{\sqrt2}\,\text{m}\)
ⓑ. \(\dfrac12\,\text{m},\;\dfrac{1}{\sqrt2}\,\text{m}\)
ⓒ. \(1\,\text{m},\;\dfrac12\,\text{m}\)
ⓓ. \(\dfrac{1}{\sqrt2}\,\text{m},\;\dfrac12\,\text{m}\)
Correct Answer: \(\dfrac{1}{\sqrt2}\,\text{m},\;\dfrac12\,\text{m}\)
Explanation: The two initial velocity vectors are perpendicular.
Their total momentum is
\[
\vec p=6\hat i+8\hat j,
\]
with magnitude
\[
p=10\,\text{kg m s}^{-1}
\]
The initial total kinetic energy is
\[
K_i=\frac12(6^2)+\frac12(8^2)=50\,\text{J}
\]
\(\textbf{Trial E — one particle stops elastically:}\)
If particle 1 stops, particle 2 must carry the entire momentum:
\[
v_E=10\,\text{m s}^{-1}
\]
Its kinetic energy is therefore
\[
K_E=\frac12(1)(10^2)=50\,\text{J},
\]
consistent with elasticity.
The spring condition is
\[
\frac12(200)x_E^2=50
\]
Hence
\[
x_E=\frac1{\sqrt2}\,\text{m}
\]
\(\textbf{Trial P — perfectly inelastic collision:}\)
The combined mass is \(2\,\text{kg}\), so
\[
V_P=\frac{10}{2}=5\,\text{m s}^{-1}
\]
Its kinetic energy is
\[
K_P=\frac12(2)(5^2)=25\,\text{J}
\]
Therefore
\[
\frac12(200)x_P^2=25
\]
Hence
\[
x_P=\frac12\,\text{m}
\]
\(\textbf{Final answer:}\)
\[
\boxed{x_E=\frac1{\sqrt2}\,\text{m},\qquad x_P=\frac12\,\text{m}}
\]
606. A particle starts from rest and moves in a straight line with constant acceleration \(a\). Let \(P_t\) and \(P_{2t}\) be the instantaneous powers delivered by the net force at times \(t\) and \(2t\), respectively. Let \(W_1\) be the net work done from \(0\) to \(t\), and \(W_2\) the net work done from \(t\) to \(2t\). Which relation is correct?
ⓐ. \(P_t:P_{2t}=1:2\) and \(W_1:W_2=1:2\)
ⓑ. \(P_t:P_{2t}=1:4\) and \(W_1:W_2=1:3\)
ⓒ. \(P_t:P_{2t}=1:2\) and \(W_1:W_2=1:3\)
ⓓ. \(P_t:P_{2t}=1:1\) and \(W_1:W_2=1:3\)
Correct Answer: \(P_t:P_{2t}=1:2\) and \(W_1:W_2=1:3\)
Explanation: The acceleration is constant, so the net force is
\[
F_{\text{net}}=ma
\]
Starting from rest, the speed at time \(t\) is
\[
v_t=at
\]
At time \(2t\),
\[
v_{2t}=2at
\]
\(\textbf{Instantaneous-power comparison:}\)
Because force and velocity are parallel,
\[
P=F_{\text{net}}v
\]
Hence
\[
P_t=ma(at)=ma^2t,
\]
while
\[
P_{2t}=ma(2at)=2ma^2t
\]
Therefore
\[
P_t:P_{2t}=1:2
\]
\(\textbf{Work in the two time intervals:}\)
From \(0\) to \(t\),
\[
W_1=\Delta K=\frac12m(at)^2
\]
From \(t\) to \(2t\),
\[
W_2=\frac12m[(2at)^2-(at)^2]
\]
Thus
\[
W_2=\frac32ma^2t^2=3W_1
\]
\(\textbf{Final answer:}\)
\[
\boxed{P_t:P_{2t}=1:2,\qquad W_1:W_2=1:3}
\]
607. A particle starts from rest and moves unidirectionally under an ideal source delivering constant positive mechanical power \(P\). There is no resistance. If \(x_t\) and \(x_{4t}\) are the displacements from the starting point at times \(t\) and \(4t\), while \(v_t\) and \(v_{4t}\) are the corresponding speeds, which relation is correct?
ⓐ. \(x_{4t}:x_t=4:1,\quad v_{4t}:v_t=4:1\)
ⓑ. \(x_{4t}:x_t=8:1,\quad v_{4t}:v_t=2:1\)
ⓒ. \(x_{4t}:x_t=8:1,\quad v_{4t}:v_t=4:1\)
ⓓ. \(x_{4t}:x_t=16:1,\quad v_{4t}:v_t=2:1\)
Correct Answer: \(x_{4t}:x_t=8:1,\quad v_{4t}:v_t=2:1\)
Explanation: Constant power means that the kinetic-energy gain is proportional to elapsed time:
\[
Pt=\frac12mv^2
\]
Hence
\[
v=\sqrt{\frac{2Pt}{m}}
\]
Therefore
\[
v\propto t^{1/2}
\]
Thus
\[
\frac{v_{4t}}{v_t}=\sqrt4=2
\]
\(\textbf{Displacement scaling:}\)
Since
\[
v=\frac{dx}{dt}\propto t^{1/2},
\]
integration with respect to time gives
\[
x\propto t^{3/2}
\]
More explicitly,
\[
x(t)=\frac23\sqrt{\frac{2P}{m}}\,t^{3/2}
\]
Therefore
\[
\frac{x_{4t}}{x_t}=4^{3/2}=8
\]
The speed does not increase linearly with time under constant power because the driving force decreases as speed rises.
\(\textbf{Final answer:}\)
\[
\boxed{x_{4t}:x_t=8:1,\qquad v_{4t}:v_t=2:1}
\]
608.
For a \(2\,\text{kg}\) particle moving along \(+x\), the net-force versus position graph is as follows. From \(x=0\) to \(2\,\text{m}\), the force increases linearly from \(0\) to \(8\,\text{N}\). From \(x=2\) to \(4\,\text{m}\), the force remains \(8\,\text{N}\). From \(x=4\) to \(5\,\text{m}\), it decreases linearly from \(8\,\text{N}\) to \(2\,\text{N}\). The particle has speed \(2\,\text{m s}^{-1}\) at \(x=0\).
What are its speed and the instantaneous net power at \(x=5\,\text{m}\)?
ⓐ. \(\sqrt{29}\,\text{m s}^{-1},\;2\sqrt{29}\,\text{W}\)
ⓑ. \(\sqrt{33}\,\text{m s}^{-1},\;8\sqrt{33}\,\text{W}\)
ⓒ. \(\sqrt{33}\,\text{m s}^{-1},\;2\sqrt{33}\,\text{W}\)
ⓓ. \(\sqrt{37}\,\text{m s}^{-1},\;2\sqrt{37}\,\text{W}\)
Correct Answer: \(\sqrt{33}\,\text{m s}^{-1},\;2\sqrt{33}\,\text{W}\)
Explanation: \(\textbf{Initial kinetic energy:}\)
\[
K_i=\frac12(2)(2^2)=4\,\text{J}
\]
\(\textbf{Work from }0\textbf{ to }2\,\text{m}\textbf{:}\)
The graph area is triangular:
\[
W_1=\frac12(2)(8)=8\,\text{J}
\]
\(\textbf{Work from }2\textbf{ to }4\,\text{m}\textbf{:}\)
\[
W_2=(8)(2)=16\,\text{J}
\]
\(\textbf{Work from }4\textbf{ to }5\,\text{m}\textbf{:}\)
The force changes linearly, so the signed trapezoidal area is
\[
W_3=\frac{8+2}{2}(1)=5\,\text{J}
\]
Hence
\[
W_{\text{net}}=8+16+5=29\,\text{J}
\]
By the work-energy theorem,
\[
K_f=4+29=33\,\text{J}
\]
For \(m=2\,\text{kg}\),
\[
33=\frac12(2)v^2,
\]
so
\[
v=\sqrt{33}\,\text{m s}^{-1}
\]
At \(x=5\,\text{m}\), the net force is \(2\,\text{N}\) along the velocity.
Thus
\[
P_{\text{net}}=Fv=2\sqrt{33}\,\text{W}
\]
\(\textbf{Final answer:}\)
\[
\boxed{v=\sqrt{33}\,\text{m s}^{-1},\qquad P_{\text{net}}=2\sqrt{33}\,\text{W}}
\]
609.
A \(2\,\text{kg}\) block starts from rest on a smooth horizontal surface. The mechanical power delivered to it increases linearly from \(0\) at \(t=0\) to \(24\,\text{W}\) at \(t=2\,\text{s}\), and then decreases linearly to zero at \(t=4\,\text{s}\). The driving force then ceases. The block immediately moves up a rough incline for which \(\sin\theta=3/5\), while a constant frictional force of \(4\,\text{N}\) opposes the motion. Take \(g=10\,\text{m s}^{-2}\).
What are the total energy supplied by the drive, the average power during the first \(4\,\text{s}\), and the distance travelled up the incline before stopping?
ⓐ. \(24\,\text{J},\;6\,\text{W},\;1.5\,\text{m}\)
ⓑ. \(48\,\text{J},\;12\,\text{W},\;3\,\text{m}\)
ⓒ. \(48\,\text{J},\;24\,\text{W},\;2\,\text{m}\)
ⓓ. \(96\,\text{J},\;24\,\text{W},\;6\,\text{m}\)
Correct Answer: \(48\,\text{J},\;12\,\text{W},\;3\,\text{m}\)
Explanation: \(\textbf{Energy from the power-time graph:}\)
The transferred energy is the area under the \(P-t\) graph.
The graph is a triangle of base \(4\,\text{s}\) and height \(24\,\text{W}\):
\[
W=\frac12(4)(24)=48\,\text{J}
\]
The block starts from rest on a smooth horizontal surface, so this becomes its kinetic energy before entering the incline.
\(\textbf{Average power:}\)
\[
P_{\text{avg}}=\frac{W}{\Delta t}
\]
Thus
\[
P_{\text{avg}}=\frac{48}{4}=12\,\text{W}
\]
\(\textbf{Opposing forces on the incline:}\)
The component of weight down the incline is
\[
mg\sin\theta=(2)(10)\left(\frac35\right)=12\,\text{N}
\]
Including friction,
\[
F_{\text{opp}}=12+4=16\,\text{N}
\]
At the stopping point, the total negative work equals the initial \(48\,\text{J}\) kinetic energy:
\[
16s=48
\]
Hence
\[
s=3\,\text{m}
\]
\(\textbf{Final answer:}\)
\[
\boxed{W=48\,\text{J},\qquad P_{\text{avg}}=12\,\text{W},\qquad s=3\,\text{m}}
\]
610.
The potential energy \(U(x)\) of a one-dimensional conservative interaction is piecewise linear. From \(x=0\) to \(3\,\text{m}\), \(U\) decreases linearly from \(12\,\text{J}\) to \(0\). From \(x=3\) to \(5\,\text{m}\), \(U\) increases linearly from \(0\) to \(6\,\text{J}\). A constant \(1\,\text{N}\) frictional force opposes motion toward \(+x\).
At one instant the particle is at \(x=1\,\text{m}\) moving toward \(+x\) at \(3\,\text{m s}^{-1}\). At another instant it is at \(x=4\,\text{m}\) moving toward \(+x\) at \(2\,\text{m s}^{-1}\). What are the net instantaneous powers at these two instants?
ⓐ. \(+9\,\text{W},\;+8\,\text{W}\)
ⓑ. \(+12\,\text{W},\;-6\,\text{W}\)
ⓒ. \(+9\,\text{W},\;-8\,\text{W}\)
ⓓ. \(+15\,\text{W},\;-4\,\text{W}\)
Correct Answer: \(+9\,\text{W},\;-8\,\text{W}\)
Explanation: The conservative force is related to the slope of the potential-energy graph by
\[
F_c=-\frac{dU}{dx}
\]
\(\textbf{At }x=1\,\text{m}\textbf{:}\)
Over the first segment,
\[
\frac{dU}{dx}=\frac{0-12}{3-0}=-4\,\text{N}
\]
Hence
\[
F_c=+4\,\text{N}
\]
Friction contributes \(-1\,\text{N}\), so
\[
F_{\text{net}}=+3\,\text{N}
\]
Therefore
\[
P_{\text{net}}=F_{\text{net}}v=(3)(3)=+9\,\text{W}
\]
\(\textbf{At }x=4\,\text{m}\textbf{:}\)
On the second segment,
\[
\frac{dU}{dx}=\frac{6-0}{5-3}=+3\,\text{N}
\]
Thus
\[
F_c=-3\,\text{N}
\]
Including friction,
\[
F_{\text{net}}=-3-1=-4\,\text{N}
\]
The velocity is \(+2\,\text{m s}^{-1}\), so
\[
P_{\text{net}}=(-4)(2)=-8\,\text{W}
\]
Positive net power means kinetic energy is increasing at the first instant; negative net power means it is decreasing at the second.
\(\textbf{Final answer:}\)
\[
\boxed{+9\,\text{W},\;-8\,\text{W}}
\]
611.
| Spring deformation \(x\) | Measured spring-force magnitude |
|---|
| \(0.10\,\text{m}\) | \(8\,\text{N}\) |
| \(0.25\,\text{m}\) | \(20\,\text{N}\) |
A \(2\,\text{kg}\) block attached to this ideal spring is released from rest on a smooth horizontal surface at \(x=+0.30\,\text{m}\). When it later passes through \(x=+0.10\,\text{m}\) while moving toward equilibrium, what are the spring constant, the block's speed and the instantaneous power delivered by the spring?
ⓐ. \(80\,\text{N m}^{-1},\;\dfrac{4}{\sqrt5}\,\text{m s}^{-1},\;\dfrac{32}{\sqrt5}\,\text{W}\)
ⓑ. \(80\,\text{N m}^{-1},\;2\,\text{m s}^{-1},\;-16\,\text{W}\)
ⓒ. \(100\,\text{N m}^{-1},\;\dfrac{4}{\sqrt5}\,\text{m s}^{-1},\;\dfrac{40}{\sqrt5}\,\text{W}\)
ⓓ. \(80\,\text{N m}^{-1},\;\dfrac{2}{\sqrt5}\,\text{m s}^{-1},\;-\dfrac{16}{\sqrt5}\,\text{W}\)
Correct Answer: \(80\,\text{N m}^{-1},\;\dfrac{4}{\sqrt5}\,\text{m s}^{-1},\;\dfrac{32}{\sqrt5}\,\text{W}\)
Explanation: \(\textbf{Infer the spring constant from the measurements:}\)
For an ideal spring,
\[
|F_s|=k|x|
\]
Using either table row,
\[
k=\frac{8}{0.10}=80\,\text{N m}^{-1}
\]
The second row confirms
\[
\frac{20}{0.25}=80\,\text{N m}^{-1}
\]
\(\textbf{Mechanical-energy conservation:}\)
Initially,
\[
E=\frac12k(0.30)^2
\]
Thus
\[
E=\frac12(80)(0.09)=3.6\,\text{J}
\]
At \(x=0.10\,\text{m}\),
\[
U_s=\frac12(80)(0.10)^2=0.4\,\text{J}
\]
Hence
\[
K=3.6-0.4=3.2\,\text{J}
\]
For \(m=2\,\text{kg}\),
\[
3.2=\frac12(2)v^2,
\]
so
\[
v=\sqrt{3.2}=\frac4{\sqrt5}\,\text{m s}^{-1}
\]
At positive \(x\), the spring force points toward negative \(x\).
The block is also moving toward negative \(x\), so spring force and velocity are parallel.
Therefore
\[
P_s=|F_s|v
\]
At \(x=0.10\,\text{m}\),
\[
P_s=(8)\left(\frac4{\sqrt5}\right)=\frac{32}{\sqrt5}\,\text{W}
\]
\(\textbf{Final answer:}\)
\[
\boxed{k=80\,\text{N m}^{-1},\quad v=\frac4{\sqrt5}\,\text{m s}^{-1},\quad P_s=\frac{32}{\sqrt5}\,\text{W}}
\]
612. Two blocks of masses \(m\) and \(3m\) are initially at rest on a smooth horizontal surface with a compressed ideal spring held between them. The spring is not attached to either block. Its stored energy is \(16\,\text{J}\). The system is released, and the blocks lose contact with the spring when it reaches its natural length. Which set correctly gives the kinetic energies of the \(m\) and \(3m\) blocks at that instant and the ratio of their speeds?
ⓐ. \(8\,\text{J},8\,\text{J};\quad v_m:v_{3m}=1:1\)
ⓑ. \(12\,\text{J},4\,\text{J};\quad v_m:v_{3m}=3:1\)
ⓒ. \(4\,\text{J},12\,\text{J};\quad v_m:v_{3m}=1:3\)
ⓓ. \(12\,\text{J},4\,\text{J};\quad v_m:v_{3m}=\sqrt3:1\)
Correct Answer: \(12\,\text{J},4\,\text{J};\quad v_m:v_{3m}=3:1\)
Explanation: The system begins from rest, so its total momentum is initially zero.
There is no external horizontal force.
Therefore the two final momenta must be equal in magnitude and opposite in direction:
\[
p_m=p_{3m}=p
\]
At natural length the spring has released all its initial \(16\,\text{J}\) of stored energy:
\[
K_m+K_{3m}=16\,\text{J}
\]
For equal momentum magnitudes,
\[
K=\frac{p^2}{2M}
\]
Thus
\[
\frac{K_m}{K_{3m}}
=
\frac{p^2/(2m)}{p^2/(6m)}
=3
\]
Hence the energies divide in the ratio
\[
K_m:K_{3m}=3:1
\]
Therefore
\[
K_m=12\,\text{J},
\qquad
K_{3m}=4\,\text{J}
\]
Momentum equality also gives
\[
mv_m=3m\,v_{3m}
\]
Thus
\[
v_m:v_{3m}=3:1
\]
The lighter block receives more kinetic energy even though the spring exerts internal forces of equal magnitude on the two blocks.
\(\textbf{Final answer:}\)
\[
\boxed{K_m=12\,\text{J},\qquad K_{3m}=4\,\text{J},\qquad v_m:v_{3m}=3:1}
\]
613.
Two identical \(2\,\text{kg}\) blocks start separately from rest at the same height \(2\,\text{m}\) above a common horizontal spring level. Both tracks have the same endpoints. On Route A, a constant \(2\,\text{N}\) frictional force acts over \(4\,\text{m}\); on Route B, the same frictional force acts over \(11\,\text{m}\). After leaving the rough portions, each block reaches an identical ideal spring of force constant \(100\,\text{N m}^{-1}\) on a smooth horizontal region. Take \(g=10\,\text{m s}^{-2}\).
What are the maximum spring compressions on Routes A and B?
ⓐ. \(0.80\,\text{m},\;0.60\,\text{m}\)
ⓑ. \(0.60\,\text{m},\;0.80\,\text{m}\)
ⓒ. \(0.80\,\text{m},\;0.40\,\text{m}\)
ⓓ. \(1.0\,\text{m},\;0.60\,\text{m}\)
Correct Answer: \(0.80\,\text{m},\;0.60\,\text{m}\)
Explanation: The gravitational potential-energy decrease is the same on both routes because the initial and final heights are the same:
\[
mgh=(2)(10)(2)=40\,\text{J}
\]
The difference between the routes comes only from the path-dependent frictional work.
\(\textbf{Route A:}\)
\[
W_{f,A}=-(2)(4)=-8\,\text{J}
\]
Hence the mechanical energy reaching the spring is
\[
E_A=40-8=32\,\text{J}
\]
At maximum compression,
\[
32=\frac12(100)x_A^2
\]
Thus
\[
50x_A^2=32,
\]
so
\[
x_A=0.80\,\text{m}
\]
\(\textbf{Route B:}\)
\[
W_{f,B}=-(2)(11)=-22\,\text{J}
\]
Therefore
\[
E_B=40-22=18\,\text{J}
\]
At maximum compression,
\[
18=\frac12(100)x_B^2
\]
Hence
\[
50x_B^2=18,
\]
giving
\[
x_B=0.60\,\text{m}
\]
Gravity contributes the same energy change on both routes, while friction distinguishes the outcomes through total path length.
\(\textbf{Final answer:}\)
\[
\boxed{x_A=0.80\,\text{m},\qquad x_B=0.60\,\text{m}}
\]
614. A \(1\,\text{kg}\) particle moves along \(+x\) with speed \(6\,\text{m s}^{-1}\) at \(x=0\). For \(x\gt 0\), the only force along the motion is
\[
F(x)=-2x\ \text{N},
\]
with \(x\) in metres. Which set gives the speed and instantaneous power at \(x=3\,\text{m}\), followed by the position where the particle first stops?
ⓐ. \(3\sqrt2\,\text{m s}^{-1},\;-18\sqrt2\,\text{W},\;3\sqrt2\,\text{m}\)
ⓑ. \(3\,\text{m s}^{-1},\;-18\,\text{W},\;6\,\text{m}\)
ⓒ. \(3\sqrt2\,\text{m s}^{-1},\;+18\sqrt2\,\text{W},\;3\sqrt2\,\text{m}\)
ⓓ. \(2\sqrt3\,\text{m s}^{-1},\;-12\sqrt3\,\text{W},\;3\,\text{m}\)
Correct Answer: \(3\sqrt2\,\text{m s}^{-1},\;-18\sqrt2\,\text{W},\;3\sqrt2\,\text{m}\)
Explanation: The initial kinetic energy is
\[
K_0=\frac12(1)(6^2)=18\,\text{J}
\]
\(\textbf{Cumulative work up to position }x\textbf{:}\)
\[
W(0\to x)=\int_0^x(-2x')\,dx'
\]
Thus
\[
W=-x^2
\]
Therefore
\[
K(x)=18-x^2
\]
\(\textbf{At }x=3\,\text{m}\textbf{:}\)
\[
K=18-9=9\,\text{J}
\]
Hence
\[
\frac12v^2=9,
\]
so
\[
v=3\sqrt2\,\text{m s}^{-1}
\]
At this position,
\[
F=-2(3)=-6\,\text{N}
\]
Because the particle is still moving toward \(+x\),
\[
P=Fv
\]
Thus
\[
P=(-6)(3\sqrt2)=-18\sqrt2\,\text{W}
\]
\(\textbf{First stopping position:}\)
Stopping occurs when
\[
K=0
\]
Therefore
\[
18-x^2=0
\]
Hence
\[
x=3\sqrt2\,\text{m}
\]
\(\textbf{Final answer:}\)
\[
\boxed{v=3\sqrt2\,\text{m s}^{-1},\quad P=-18\sqrt2\,\text{W},\quad x_{\text{stop}}=3\sqrt2\,\text{m}}
\]
615. A \(0.20\,\text{kg}\) object is released from rest and falls vertically through \(20\,\text{m}\). During the first \(8\,\text{m}\), air resistance increases in effect so that the object's speed becomes \(10\,\text{m s}^{-1}\). It then continues downward at this constant terminal speed for the remaining \(12\,\text{m}\). Take \(g=10\,\text{m s}^{-2}\). What are the works done by air resistance during the first \(8\,\text{m}\), during the final \(12\,\text{m}\), and during the complete \(20\,\text{m}\) descent?
ⓐ. \(-16\,\text{J},\;-24\,\text{J},\;-40\,\text{J}\)
ⓑ. \(-10\,\text{J},\;-20\,\text{J},\;-30\,\text{J}\)
ⓒ. \(+6\,\text{J},\;-24\,\text{J},\;-18\,\text{J}\)
ⓓ. \(-6\,\text{J},\;-24\,\text{J},\;-30\,\text{J}\)
Correct Answer: \(-6\,\text{J},\;-24\,\text{J},\;-30\,\text{J}\)
Explanation: \(\textbf{First }8\,\text{m}\textbf{:}\)
The gravitational work is
\[
W_{g,1}=mgs
\]
Thus
\[
W_{g,1}=(0.20)(10)(8)=16\,\text{J}
\]
The kinetic-energy increase is
\[
\Delta K_1=\frac12(0.20)(10^2)-0
\]
Hence
\[
\Delta K_1=10\,\text{J}
\]
By the work-energy theorem,
\[
W_{g,1}+W_{r,1}=\Delta K_1
\]
Therefore
\[
16+W_{r,1}=10,
\]
so
\[
W_{r,1}=-6\,\text{J}
\]
\(\textbf{Final }12\,\text{m}\textbf{ at terminal speed:}\)
The kinetic energy is unchanged:
\[
\Delta K_2=0
\]
Gravity does
\[
W_{g,2}=(0.20)(10)(12)=24\,\text{J}
\]
Thus resistance must do
\[
W_{r,2}=-24\,\text{J}
\]
\(\textbf{Complete descent:}\)
\[
W_r=-6-24=-30\,\text{J}
\]
The resistive work is not simply \(-mgh\) for the whole fall because the object finishes with non-zero kinetic energy.
\(\textbf{Final answer:}\)
\[
\boxed{W_{r,1}=-6\,\text{J},\qquad W_{r,2}=-24\,\text{J},\qquad W_r=-30\,\text{J}}
\]
616. A \(1000\,\text{kg}\) car is moving uphill at \(10\,\text{m s}^{-1}\) on a road for which \(\sin\theta=0.05\). A constant \(500\,\text{N}\) resistive force opposes its motion, and at this instant the car is accelerating uphill at \(0.50\,\text{m s}^{-2}\). Take \(g=10\,\text{m s}^{-2}\). If the engine delivers the required tractive force parallel to the road, which set gives the instantaneous engine power and the acceleration that the same tractive power would produce at the same speed on a horizontal resistance-free road?
ⓐ. \(15\,\text{kW},\;1.5\,\text{m s}^{-2}\)
ⓑ. \(10\,\text{kW},\;1.0\,\text{m s}^{-2}\)
ⓒ. \(20\,\text{kW},\;2.0\,\text{m s}^{-2}\)
ⓓ. \(15\,\text{kW},\;0.5\,\text{m s}^{-2}\)
Correct Answer: \(15\,\text{kW},\;1.5\,\text{m s}^{-2}\)
Explanation: \(\textbf{Forces required during the uphill acceleration:}\)
Along the incline,
\[
F_{\text{engine}}-mg\sin\theta-f=ma
\]
Hence
\[
F_{\text{engine}}
=
ma+mg\sin\theta+f
\]
Substituting,
\[
F_{\text{engine}}
=
(1000)(0.50)+(1000)(10)(0.05)+500
\]
Thus
\[
F_{\text{engine}}=500+500+500=1500\,\text{N}
\]
\(\textbf{Instantaneous engine power:}\)
Because the tractive force is parallel to the velocity,
\[
P=F_{\text{engine}}v
\]
Therefore
\[
P=(1500)(10)=1.5\times10^4\,\text{W}=15\,\text{kW}
\]
\(\textbf{Changed road condition:}\)
On a horizontal resistance-free road at the same speed,
\[
P=Fv=mav
\]
Hence
\[
a=\frac{P}{mv}
\]
Substituting,
\[
a=\frac{15000}{(1000)(10)}=1.5\,\text{m s}^{-2}
\]
The same power can produce a larger acceleration after the gravitational and resistive power demands are removed.
\(\textbf{Final answer:}\)
\[
\boxed{P=15\,\text{kW},\qquad a=1.5\,\text{m s}^{-2}}
\]
617. A \(50\,\text{kg}\) load is lifted vertically upward through \(10\,\text{m}\) in \(5\,\text{s}\). It starts from rest and has speed \(4\,\text{m s}^{-1}\) at the end of the \(5\,\text{s}\) interval. Neglect all losses and take \(g=10\,\text{m s}^{-2}\). What is the average mechanical power supplied to the load during the interval?
ⓐ. \(1000\,\text{W}\)
ⓑ. \(1080\,\text{W}\)
ⓒ. \(1160\,\text{W}\)
ⓓ. \(1200\,\text{W}\)
Correct Answer: \(1080\,\text{W}\)
Explanation: The supplied work must account for both the increase in gravitational potential energy and the increase in kinetic energy.
\(\textbf{Gravitational-energy increase:}\)
\[
\Delta U=mgh
\]
Thus
\[
\Delta U=(50)(10)(10)=5000\,\text{J}
\]
\(\textbf{Kinetic-energy increase:}\)
The load starts from rest, so
\[
\Delta K=\frac12mv^2
\]
Hence
\[
\Delta K=\frac12(50)(4^2)=400\,\text{J}
\]
Therefore the total mechanical work supplied is
\[
W=\Delta U+\Delta K
\]
Thus
\[
W=5000+400=5400\,\text{J}
\]
\(\textbf{Average power:}\)
\[
P_{\text{avg}}=\frac{W}{\Delta t}
\]
Therefore
\[
P_{\text{avg}}=\frac{5400}{5}=1080\,\text{W}
\]
Using only \(mgh/t\) would miss the kinetic-energy gained by the load.
\(\textbf{Final answer:}\)
\[
\boxed{P_{\text{avg}}=1080\,\text{W}}
\]
618. A particle moves from point A to point B under only a conservative force. With one choice of potential-energy zero,
\[
K_A=5\,\text{J},\qquad U_A=20\,\text{J},\qquad U_B=8\,\text{J}.
\]
A second observer chooses a different zero of potential energy such that every potential-energy value is increased by \(30\,\text{J}\). Which statement is correct?
ⓐ. The second observer obtains \(K_B=47\,\text{J}\) because \(U_B\) becomes \(38\,\text{J}\)
ⓑ. Both observers obtain \(K_B=17\,\text{J}\) and work by the conservative force \(=12\,\text{J}\)
ⓒ. The second observer obtains work \(=-18\,\text{J}\) because the absolute value of \(U_B\) changes
ⓓ. Mechanical-energy conservation is valid only for the first choice of potential-energy zero
Correct Answer: Both observers obtain \(K_B=17\,\text{J}\) and work by the conservative force \(=12\,\text{J}\)
Explanation: With the first reference choice, the mechanical energy at A is
\[
E=K_A+U_A.
\]
Therefore
\[
E=5+20=25\,\text{J}
\]
At B,
\[
K_B+U_B=25\,\text{J}
\]
Hence
\[
K_B=25-8=17\,\text{J}
\]
The work done by the conservative force is
\[
W_c=-\Delta U.
\]
Thus
\[
W_c=U_A-U_B=20-8=12\,\text{J}
\]
\(\textbf{Shift of potential-energy zero:}\)
The second observer assigns
\[
U_A'=50\,\text{J},
\qquad
U_B'=38\,\text{J}
\]
The new total mechanical energy is
\[
E'=K_A+U_A'=55\,\text{J}
\]
Therefore
\[
K_B'=55-38=17\,\text{J}
\]
Also,
\[
U_A'-U_B'=50-38=12\,\text{J}
\]
Adding the same constant to every potential-energy value changes neither potential-energy differences nor the force, work or kinetic-energy predictions.
\(\textbf{Final answer:}\)
\[
\boxed{K_B=17\,\text{J},\qquad W_c=12\,\text{J}\text{ for both observers}}
\]
619.
A \(2\,\text{kg}\) block can move along a smooth incline for which \(\sin\theta=1/2\). It is attached to an ideal spring of force constant \(200\,\text{N m}^{-1}\), with the spring lying along the incline. Let \(x=0\) correspond to the spring's natural length, and let positive \(x\) be uphill. The block is released from rest at \(x=-0.20\,\text{m}\), where the spring is compressed. Take \(g=10\,\text{m s}^{-2}\).
What are the block's speed as it first passes \(x=0\) and its maximum positive extension of the spring?
ⓐ. \(\sqrt2\,\text{m s}^{-1},\;0.10\,\text{m}\)
ⓑ. \(2\,\text{m s}^{-1},\;0.20\,\text{m}\)
ⓒ. \(\sqrt2\,\text{m s}^{-1},\;0.20\,\text{m}\)
ⓓ. \(1\,\text{m s}^{-1},\;0.10\,\text{m}\)
Correct Answer: \(\sqrt2\,\text{m s}^{-1},\;0.10\,\text{m}\)
Explanation: Both gravity and the spring are conservative, so total mechanical energy is conserved.
Choose gravitational potential energy to be zero at \(x=0\).
\(\textbf{Initial mechanical energy at }x=-0.20\,\text{m}\textbf{:}\)
The spring energy is
\[
U_{s,i}=\frac12kx^2
\]
Thus
\[
U_{s,i}=\frac12(200)(0.20)^2=4\,\text{J}
\]
The vertical coordinate changes with distance along the incline:
\[
h=x\sin\theta
\]
Hence initially
\[
U_{g,i}=mg(-0.20)\left(\frac12\right)
\]
Therefore
\[
U_{g,i}=(2)(10)(-0.10)=-2\,\text{J}
\]
The block starts from rest, so
\[
E=4-2=2\,\text{J}
\]
\(\textbf{Speed at the spring's natural length:}\)
At \(x=0\),
\[
U_s=0,\qquad U_g=0
\]
Therefore
\[
K=2\,\text{J}
\]
Thus
\[
\frac12(2)v^2=2,
\]
giving
\[
v=\sqrt2\,\text{m s}^{-1}
\]
\(\textbf{Maximum uphill extension:}\)
At the turning point \(x\gt 0\), the speed is zero:
\[
\frac12kx^2+mgx\sin\theta=2
\]
Substituting,
\[
100x^2+10x=2
\]
Hence
\[
100x^2+10x-2=0
\]
The physical positive root is
\[
x=0.10\,\text{m}
\]
\(\textbf{Final answer:}\)
\[
\boxed{v=\sqrt2\,\text{m s}^{-1},\qquad x_{\max}=0.10\,\text{m}}
\]
620.
A \(2\,\text{kg}\) block moves from point A at height \(5\,\text{m}\) to point B at height \(1\,\text{m}\) along a rough track of length \(10\,\text{m}\). Its speed is \(2\,\text{m s}^{-1}\) at A and \(8\,\text{m s}^{-1}\) at B. The frictional force has constant magnitude throughout the motion. Take \(g=10\,\text{m s}^{-2}\).
What are the work done by friction and the frictional-force magnitude?
ⓐ. \(+20\,\text{J},\;2\,\text{N}\)
ⓑ. \(-40\,\text{J},\;4\,\text{N}\)
ⓒ. \(-20\,\text{J},\;2\,\text{N}\)
ⓓ. \(-60\,\text{J},\;6\,\text{N}\)
Correct Answer: \(-20\,\text{J},\;2\,\text{N}\)
Explanation: For a system with gravity treated through potential energy, the work of friction equals the change in mechanical energy:
\[
W_f=\Delta(K+U)
\]
\(\textbf{Mechanical energy at A:}\)
\[
K_A=\frac12(2)(2^2)=4\,\text{J}
\]
\[
U_A=(2)(10)(5)=100\,\text{J}
\]
Hence
\[
E_A=104\,\text{J}
\]
\(\textbf{Mechanical energy at B:}\)
\[
K_B=\frac12(2)(8^2)=64\,\text{J}
\]
\[
U_B=(2)(10)(1)=20\,\text{J}
\]
Thus
\[
E_B=84\,\text{J}
\]
Therefore
\[
W_f=E_B-E_A
\]
Hence
\[
W_f=84-104=-20\,\text{J}
\]
The negative sign is consistent with friction removing mechanical energy.
Because the friction magnitude is constant,
\[
W_f=-fs
\]
Therefore
\[
-20=-f(10)
\]
Hence
\[
f=2\,\text{N}
\]
\(\textbf{Final answer:}\)
\[
\boxed{W_f=-20\,\text{J},\qquad f=2\,\text{N}}
\]
