101. Two identical balls approach a catcher with the same speed \(v\). Ball \(P\) is brought to rest in time \(\Delta t\). Ball \(Q\) is caught in such a way that it rebounds backward with speed \(v/2\), and its contact time is \(3\Delta t\). What is the ratio of the magnitudes of the average forces on \(Q\) and \(P\), \(F_Q:F_P\)?
ⓐ. \(3:2\)
ⓑ. \(2:1\)
ⓒ. \(1:2\)
ⓓ. \(1:3\)
Correct Answer: \(1:2\)
Explanation: Let the common mass be \(m\), and take the initial direction of motion as positive. For ball \(P\),
\[
p_i=mv,\qquad p_f=0.
\]
Thus,
\[
|\Delta p_P|=mv.
\]
Its average-force magnitude is
\[
F_P=\frac{mv}{\Delta t}.
\]
For ball \(Q\), the final velocity is \(-v/2\), so
\[
\Delta p_Q=m\left(-\frac v2-v\right)
=-\frac{3mv}{2}.
\]
Hence,
\[
|\Delta p_Q|=\frac{3mv}{2}.
\]
Its interaction time is \(3\Delta t\), so
\[
F_Q=\frac{(3mv/2)}{3\Delta t}
=\frac{mv}{2\Delta t}.
\]
Therefore,
\[
\frac{F_Q}{F_P}=\frac12.
\]
The rebound produces a larger momentum change than merely stopping the ball, but the three-times-longer contact interval more than compensates for that increase. Hence \(F_Q:F_P=1:2\).
102. The net force on a particle varies with time as described below.
From \(0\) to \(2\,\text{s}\), the force is constant at \(+6\,\text{N}\). From \(2\) to \(4\,\text{s}\), it decreases linearly from \(+6\,\text{N}\) to zero. From \(4\) to \(6\,\text{s}\), it is constant at \(-5\,\text{N}\). Positive force is eastward. The particle initially has momentum \(6\,\text{kg m s}^{-1}\) westward.
What is the particle's momentum at \(t=6\,\text{s}\)?
ⓐ. \(8\,\text{kg m s}^{-1}\) eastward
ⓑ. \(2\,\text{kg m s}^{-1}\) eastward
ⓒ. \(2\,\text{kg m s}^{-1}\) westward
ⓓ. \(14\,\text{kg m s}^{-1}\) eastward
Correct Answer: \(2\,\text{kg m s}^{-1}\) eastward
Explanation: Take east as positive, so the initial momentum is
\[
p_i=-6\,\text{kg m s}^{-1}.
\]
The impulse equals the signed area under the force-time graph. From \(0\) to \(2\,\text{s}\),
\[
J_1=(6)(2)=12\,\text{N s}.
\]
From \(2\) to \(4\,\text{s}\), the region is a triangle:
\[
J_2=\frac12(2)(6)=6\,\text{N s}.
\]
From \(4\) to \(6\,\text{s}\),
\[
J_3=(-5)(2)=-10\,\text{N s}.
\]
Therefore the net impulse is
\[
J_{\text{net}}=12+6-10=8\,\text{N s}.
\]
Using
\[
p_f=p_i+J_{\text{net}},
\]
we get
\[
p_f=-6+8=+2\,\text{kg m s}^{-1}.
\]
The positive result means eastward. The particle therefore reverses its momentum direction during the complete force history and ends with momentum \(2\,\text{kg m s}^{-1}\) eastward.
103. The net force acting along the \(x\)-axis on a particle varies as follows.
From \(t=0\) to \(2\,\text{s}\), the force increases linearly from \(+2\,\text{N}\) to \(+6\,\text{N}\). From \(t=2\) to \(4\,\text{s}\), the force is constant at \(-3\,\text{N}\). The particle initially has momentum \(+1\,\text{kg m s}^{-1}\).
Which option correctly gives the net impulse, final momentum and average net force over the complete \(4\,\text{s}\) interval?
ⓐ. \(2\,\text{N s}\), \(3\,\text{kg m s}^{-1}\), \(0.5\,\text{N}\) in the positive direction
ⓑ. \(8\,\text{N s}\), \(9\,\text{kg m s}^{-1}\), \(2\,\text{N}\) in the positive direction
ⓒ. \(14\,\text{N s}\), \(15\,\text{kg m s}^{-1}\), \(3.5\,\text{N}\) in the positive direction
ⓓ. \(-2\,\text{N s}\), \(-1\,\text{kg m s}^{-1}\), \(0.5\,\text{N}\) in the negative direction
Correct Answer: \(2\,\text{N s}\), \(3\,\text{kg m s}^{-1}\), \(0.5\,\text{N}\) in the positive direction
Explanation: \( \textbf{First interval:} \) The force varies linearly from \(2\) to \(6\,\text{N}\), so its average value during these \(2\,\text{s}\) is
\[
F_{\text{avg,1}}=\frac{2+6}{2}=4\,\text{N}.
\]
Hence,
\[
J_1=4(2)=8\,\text{N s}.
\]
\( \textbf{Second interval:} \)
\[
J_2=(-3)(2)=-6\,\text{N s}.
\]
Therefore,
\[
J_{\text{net}}=8-6=2\,\text{N s}.
\]
The final momentum is
\[
p_f=p_i+J_{\text{net}}=1+2=3\,\text{kg m s}^{-1}.
\]
The average net force over the entire \(4\,\text{s}\) interval is
\[
F_{\text{avg}}=\frac{J_{\text{net}}}{4}
=\frac24
=0.5\,\text{N}.
\]
It points in the positive direction because the signed net impulse is positive. Geometric areas below the time axis must be subtracted rather than added as positive areas.
104. A net force acts on a particle during the interval \(0\) to \(6\,\text{s}\). The positive area between the \(F-t\) graph and the time axis is \(18\,\text{N s}\), while the negative area has magnitude \(18\,\text{N s}\). The force is not zero throughout the interval. Consider the following statements.
Statement I: The net impulse over the \(6\,\text{s}\) interval is zero.
Statement II: The particle's final momentum equals its initial momentum.
Statement III: The average net force over the complete interval is zero.
Statement IV: The particle's momentum must have remained constant at every instant during the interval.
Which option is correct?
ⓐ. I and IV only
ⓑ. II and III only
ⓒ. I, II and III only
ⓓ. I, II, III and IV
Correct Answer: I, II and III only
Explanation: Impulse is the signed area under the force-time graph. The positive contribution is \(+18\,\text{N s}\), while the negative contribution is \(-18\,\text{N s}\). Therefore,
\[
J_{\text{net}}=18-18=0.
\]
Statement I is correct. Since
\[
\Delta\vec p=\vec J_{\text{net}},
\]
the total momentum change over the complete interval is zero, so the final momentum equals the initial momentum. Statement II is therefore correct. The average net force is
\[
F_{\text{avg}}=\frac{J_{\text{net}}}{\Delta t}=0,
\]
so Statement III is also correct. However, the instantaneous force was explicitly non-zero during portions of the interval. During those portions the momentum was changing, even though the later momentum changes exactly cancelled the earlier ones. Statement IV is therefore incorrect. Returning to the original momentum after an interval is not the same as having constant momentum throughout that interval.
105. A \(2\,\text{kg}\) particle initially moves with velocity
\[
\vec v_i=(3\hat i-2\hat j)\,\text{m s}^{-1}.
\]
A \(5\,\text{N}\) force acts northward for \(2\,\text{s}\). Immediately afterward, a \(3\,\text{N}\) force acts westward for \(1\,\text{s}\). No other impulse acts during these intervals. What is the particle's final speed?
ⓐ. \(\dfrac{3\sqrt2}{2}\,\text{m s}^{-1}\)
ⓑ. \(3\sqrt5\,\text{m s}^{-1}\)
ⓒ. \(\sqrt5\,\text{m s}^{-1}\)
ⓓ. \(\dfrac{3\sqrt5}{2}\,\text{m s}^{-1}\)
Correct Answer: \(\dfrac{3\sqrt5}{2}\,\text{m s}^{-1}\)
Explanation: \( \textbf{Initial momentum:} \)
\[
\vec p_i=m\vec v_i
=2(3\hat i-2\hat j)
=(6\hat i-4\hat j)\,\text{kg m s}^{-1}.
\]
The first impulse is northward:
\[
\vec J_1=(5)(2)\hat j=10\hat j\,\text{N s}.
\]
Thus after the first interval,
\[
\vec p_1=6\hat i+6\hat j.
\]
The second impulse is westward:
\[
\vec J_2=-3\hat i\,\text{N s}.
\]
Hence the final momentum is
\[
\vec p_f=(6-3)\hat i+6\hat j
=3\hat i+6\hat j.
\]
Its magnitude is
\[
p_f=\sqrt{3^2+6^2}
=3\sqrt5\,\text{kg m s}^{-1}.
\]
Since \(m=2\,\text{kg}\),
\[
v_f=\frac{p_f}{m}
=\frac{3\sqrt5}{2}\,\text{m s}^{-1}.
\]
The two impulses must be added vectorially and in sequence to the initial momentum rather than treated merely through their scalar magnitudes.
106. A \(0.50\,\text{kg}\) ball moves eastward at \(8\,\text{m s}^{-1}\). A westward force pulse has a triangular \(F-t\) graph: the force starts from zero, rises to a peak magnitude of \(20\,\text{N}\), and returns to zero after time \(T\). The pulse brings the ball exactly to rest. In a second trial, an identical ball has the same initial velocity, but the pulse has the same duration and triangular shape with only half the peak force. Which option is correct?
ⓐ. \(T=0.20\,\text{s}\); second ball stops
ⓑ. \(T=0.40\,\text{s}\); second ball continues eastward at \(4\,\text{m s}^{-1}\)
ⓒ. \(T=0.40\,\text{s}\); second ball rebounds westward at \(4\,\text{m s}^{-1}\)
ⓓ. \(T=0.80\,\text{s}\); second ball continues eastward at \(2\,\text{m s}^{-1}\)
Correct Answer: \(T=0.40\,\text{s}\); second ball continues eastward at \(4\,\text{m s}^{-1}\)
Explanation: The initial momentum is
\[
p_i=mv=0.50(8)=4\,\text{kg m s}^{-1}
\]
eastward. To bring the ball exactly to rest, the first pulse must supply an impulse of magnitude \(4\,\text{N s}\) westward. The impulse equals the area of the triangular force-time graph:
\[
J=\frac12(20)T.
\]
Therefore,
\[
10T=4,
\]
so
\[
T=0.40\,\text{s}.
\]
In the second trial, the peak force is halved to \(10\,\text{N}\), while the duration and triangular shape remain unchanged. Its impulse magnitude is therefore
\[
J_2=\frac12(10)(0.40)=2\,\text{N s}
\]
westward. Taking east as positive,
\[
p_{f2}=4-2=2\,\text{kg m s}^{-1}.
\]
Hence,
\[
v_{f2}=\frac{2}{0.50}=4\,\text{m s}^{-1}
\]
eastward. Halving the peak force in an otherwise identical triangular pulse halves the impulse, so the second ball is slowed but not stopped.
107. Ball \(P\) of mass \(1\,\text{kg}\) changes velocity from \(6\,\text{m s}^{-1}\) east to \(2\,\text{m s}^{-1}\) west in \(0.20\,\text{s}\). Ball \(Q\) of mass \(2\,\text{kg}\) changes velocity from \(3\,\text{m s}^{-1}\) east to rest in \(0.30\,\text{s}\). What is the ratio of the magnitudes of their average net forces, \(F_P:F_Q\)?
ⓐ. \(2:1\)
ⓑ. \(4:3\)
ⓒ. \(3:2\)
ⓓ. \(1:2\)
Correct Answer: \(2:1\)
Explanation: Take east as positive. For ball \(P\),
\[
\Delta p_P
=1(-2-6)
=-8\,\text{kg m s}^{-1}.
\]
Therefore,
\[
F_P=\frac{|\Delta p_P|}{\Delta t_P}
=\frac{8}{0.20}
=40\,\text{N}.
\]
For ball \(Q\),
\[
\Delta p_Q
=2(0-3)
=-6\,\text{kg m s}^{-1}.
\]
Hence,
\[
F_Q=\frac{6}{0.30}
=20\,\text{N}.
\]
Thus,
\[
F_P:F_Q=40:20=2:1.
\]
Although \(Q\) has the larger mass, it does not experience the larger average force in this comparison. Average force depends on the momentum change per unit time, not on mass alone. The reversal of \(P\)'s motion makes its momentum change particularly large.
108. A \(2\,\text{kg}\) particle initially has momentum \(3\,\text{kg m s}^{-1}\) westward. The net horizontal force then follows the table below.
| Time interval | Net force |
|---|
| \(0\) to \(1\,\text{s}\) | \(4\,\text{N}\) east |
| \(1\) to \(3\,\text{s}\) | \(2\,\text{N}\) west |
| \(3\) to \(4\,\text{s}\) | \(3\,\text{N}\) east |
Which statement is correct at \(t=4\,\text{s}\)?
ⓐ. The particle has momentum \(3\,\text{kg m s}^{-1}\) eastward and speed \(1.5\,\text{m s}^{-1}\).
ⓑ. The particle has momentum \(4\,\text{kg m s}^{-1}\) westward and speed \(2\,\text{m s}^{-1}\).
ⓒ. The particle has momentum \(1\,\text{kg m s}^{-1}\) eastward and speed \(0.5\,\text{m s}^{-1}\).
ⓓ. The particle is momentarily at rest, and the average net force over the \(4\,\text{s}\) interval is \(0.75\,\text{N}\) eastward.
Correct Answer: The particle is momentarily at rest, and the average net force over the \(4\,\text{s}\) interval is \(0.75\,\text{N}\) eastward.
Explanation: Take east as positive, so
\[
p_i=-3\,\text{kg m s}^{-1}.
\]
The first impulse is
\[
J_1=4(1)=+4\,\text{N s}.
\]
The second impulse is
\[
J_2=(-2)(2)=-4\,\text{N s}.
\]
The third impulse is
\[
J_3=3(1)=+3\,\text{N s}.
\]
Thus the total impulse is
\[
J_{\text{net}}=4-4+3=3\,\text{N s}.
\]
Therefore,
\[
p_f=p_i+J_{\text{net}}=-3+3=0.
\]
The particle is momentarily at rest at \(t=4\,\text{s}\). The average net force over the full interval is
\[
F_{\text{avg}}=\frac{J_{\text{net}}}{4}
=\frac34
=0.75\,\text{N}
\]
eastward. A zero final momentum does not imply that the force or momentum was zero throughout the preceding interval.
109. A particle initially has momentum
\[
\vec p_i=(-3\hat i+4\hat j)\,\text{kg m s}^{-1}.
\]
A constant resultant force
\[
\vec F=(6\hat i-8\hat j)\,\text{N}
\]
acts for \(0.50\,\text{s}\) and is then removed completely. Which statement is correct immediately after removal?
ⓐ. The momentum is \(5\,\text{kg m s}^{-1}\) north-east.
ⓑ. The momentum is \(5\,\text{kg m s}^{-1}\) south-east.
ⓒ. The particle is at rest and remains at rest afterward if no other resultant force acts.
ⓓ. The particle is at rest only instantaneously and then moves in the direction of the previous force.
Correct Answer: The particle is at rest and remains at rest afterward if no other resultant force acts.
Explanation: The impulse delivered by the constant force is
\[
\vec J=\vec F\Delta t.
\]
Therefore,
\[
\vec J=(6\hat i-8\hat j)(0.50)
=(3\hat i-4\hat j)\,\text{N s}.
\]
Since \(\text{N s}\) is equivalent to \(\text{kg m s}^{-1}\),
\[
\vec p_f=\vec p_i+\vec J.
\]
Thus,
\[
\vec p_f=(-3\hat i+4\hat j)+(3\hat i-4\hat j)=0.
\]
The force pulse has exactly cancelled the initial momentum vector. At the moment the force is removed, the particle therefore has zero velocity for non-zero mass. The problem then states that no other resultant force acts. Hence the acceleration after removal is zero and the zero velocity persists. The direction of the earlier force is not stored in the particle after the interaction has ended. Therefore the particle remains at rest.
110. The one-dimensional momentum of a particle varies with time as described below.
From \(t=0\) to \(2\,\text{s}\), momentum increases linearly from \(-4\) to \(+2\,\text{kg m s}^{-1}\). From \(t=2\) to \(5\,\text{s}\), momentum remains constant at \(+2\,\text{kg m s}^{-1}\). From \(t=5\) to \(6\,\text{s}\), momentum decreases linearly from \(+2\) to \(-4\,\text{kg m s}^{-1}\).
Which statement is correct?
ⓐ. First interval: zero force; middle interval: zero force; final interval: twice the first-interval force in the opposite direction.
ⓑ. First interval: non-zero force; middle interval: zero force despite non-zero momentum; final force has twice the first magnitude and opposite direction.
ⓒ. First interval: non-zero force; middle interval: non-zero force because momentum is non-zero; final force equals the first in magnitude.
ⓓ. First interval: non-zero force; middle interval: zero force; final force equals the first in magnitude because both momentum changes are \(6\,\text{kg m s}^{-1}\).
Correct Answer: First interval: non-zero force; middle interval: zero force despite non-zero momentum; final force has twice the first magnitude and opposite direction.
Explanation: On a momentum-time graph, the slope gives the net force:
\[
F=\frac{dp}{dt}.
\]
During the first interval,
\[
F_1=\frac{2-(-4)}{2-0}
=\frac62
=+3\,\text{N}.
\]
During the middle interval, momentum is constant, so the graph is horizontal and
\[
F_2=0.
\]
The particle can therefore possess non-zero momentum while experiencing zero net force. During the final interval,
\[
F_3=\frac{-4-2}{6-5}
=-6\,\text{N}.
\]
Thus the final force has twice the magnitude of the first force and points in the opposite direction. The fact that momentum passes through zero in the first interval does not make the force zero there; the slope remains positive. Hence option B correctly interprets both graph height and graph slope.
111. The momentum of a \(0.50\,\text{kg}\) particle moving along the \(x\)-axis is
\[
p_x=t^2-4t+3
\]
in SI units. Which statement about the particle at \(t=1\,\text{s}\) and \(t=3\,\text{s}\) is correct?
ⓐ. At both \(1\,\text{s}\) and \(3\,\text{s}\): \(p_x=0\) and \(a_x=0\).
ⓑ. At both instants \(p_x\ne0\); the acceleration reverses from westward to eastward.
ⓒ. At both instants \(p_x=0\); \(a_x=4\,\text{m s}^{-2}\) westward at \(1\,\text{s}\) and eastward at \(3\,\text{s}\).
ⓓ. At both instants \(p_x=0\); therefore \(a_x=0\) at both instants.
Correct Answer: At both instants \(p_x=0\); \(a_x=4\,\text{m s}^{-2}\) westward at \(1\,\text{s}\) and eastward at \(3\,\text{s}\).
Explanation: First evaluate the momentum at \(t=1\,\text{s}\):
\[
p_x(1)=1-4+3=0.
\]
At \(t=3\,\text{s}\),
\[
p_x(3)=9-12+3=0.
\]
Thus the particle is instantaneously at rest at both times. The net force is obtained from the rate of change of momentum:
\[
F_x=\frac{dp_x}{dt}=2t-4.
\]
At \(t=1\,\text{s}\),
\[
F_x=2-4=-2\,\text{N}.
\]
Hence,
\[
a_x=\frac{-2}{0.50}=-4\,\text{m s}^{-2},
\]
which is westward. At \(t=3\,\text{s}\),
\[
F_x=6-4=+2\,\text{N},
\]
so
\[
a_x=\frac{2}{0.50}=4\,\text{m s}^{-2}
\]
eastward. Therefore zero instantaneous momentum does not imply zero force or acceleration.
112. A particle moves along the \(x\)-axis. During Stage I, its momentum changes from \(+5\,\text{kg m s}^{-1}\) to \(-1\,\text{kg m s}^{-1}\) in \(0.20\,\text{s}\). During Stage II, it changes from \(-1\,\text{kg m s}^{-1}\) to \(+3\,\text{kg m s}^{-1}\) in \(0.40\,\text{s}\). Which option correctly gives the average net forces during Stage I, Stage II and the complete \(0.60\,\text{s}\) interval?
ⓐ. \(30\,\text{N}\) west, \(10\,\text{N}\) east, and \(\dfrac{10}{3}\,\text{N}\) west
ⓑ. \(20\,\text{N}\) west, \(10\,\text{N}\) east, and zero
ⓒ. \(30\,\text{N}\) east, \(10\,\text{N}\) west, and \(\dfrac{10}{3}\,\text{N}\) east
ⓓ. \(6\,\text{N}\) west, \(4\,\text{N}\) east, and \(2\,\text{N}\) west
Correct Answer: \(30\,\text{N}\) west, \(10\,\text{N}\) east, and \(\dfrac{10}{3}\,\text{N}\) west
Explanation: Take east as positive. During Stage I,
\[
\Delta p_1=-1-(+5)=-6\,\text{kg m s}^{-1}.
\]
Thus,
\[
F_{\text{avg,1}}=\frac{-6}{0.20}=-30\,\text{N},
\]
which means \(30\,\text{N}\) westward. During Stage II,
\[
\Delta p_2=+3-(-1)=+4\,\text{kg m s}^{-1}.
\]
Therefore,
\[
F_{\text{avg,2}}=\frac{4}{0.40}=10\,\text{N}
\]
eastward. For the complete interval, use final momentum minus initial momentum directly:
\[
\Delta p_{\text{total}}=3-5=-2\,\text{kg m s}^{-1}.
\]
The total time is
\[
0.20+0.40=0.60\,\text{s}.
\]
Hence,
\[
F_{\text{avg,total}}=\frac{-2}{0.60}
=-\frac{10}{3}\,\text{N}.
\]
Thus the overall average force is \(10/3\,\text{N}\) westward. The complete-interval average is not obtained by simply taking the arithmetic mean of the two stage-average forces because the stage durations differ.
113. Two identical particles initially move with the same speed \(v\) towards the east. During a time interval \(\Delta t\), particle \(P\) continues to move east with unchanged velocity, whereas particle \(Q\) emerges with the same speed \(v\) towards the north. If each particle has mass \(m\), which statement correctly compares their average net forces during the interval?
ⓐ. Both have zero average net force because their speeds are unchanged.
ⓑ. \(P\) has average force \(mv/\Delta t\) eastward, while \(Q\) has average force \(mv/\Delta t\) northward.
ⓒ. \(P\) has zero average net force; \(Q\) has average-force magnitude \(\sqrt2\,mv/\Delta t\), directed north-west.
ⓓ. \(P\) has zero average net force; \(Q\) has average-force magnitude \(2mv/\Delta t\), directed north-east.
Correct Answer: \(P\) has zero average net force; \(Q\) has average-force magnitude \(\sqrt2\,mv/\Delta t\), directed north-west.
Explanation: Particle \(P\) has the same initial and final velocity vector. Therefore its momentum does not change:
\[
\Delta\vec p_P=0.
\]
Hence its average net force is
\[
\vec F_{\text{avg},P}=0.
\]
For particle \(Q\), the initial momentum is
\[
\vec p_i=mv\,\hat i,
\]
while the final momentum is
\[
\vec p_f=mv\,\hat j.
\]
Therefore,
\[
\Delta\vec p_Q
=mv(\hat j-\hat i).
\]
Its magnitude is
\[
|\Delta\vec p_Q|
=mv\sqrt{1^2+1^2}
=\sqrt2\,mv.
\]
The negative \(x\)-component and positive \(y\)-component make the momentum change point north-west. Hence,
\[
F_{\text{avg},Q}
=\frac{\sqrt2\,mv}{\Delta t}.
\]
This result demonstrates that constant speed does not imply zero force when the direction of momentum changes. Particle \(Q\)'s force is associated with changing the direction of its momentum vector rather than its magnitude.
114. Three blocks \(A\), \(B\) and \(C\) of masses \(2\,\text{kg}\), \(3\,\text{kg}\) and \(5\,\text{kg}\), respectively, lie in contact in this order on a smooth horizontal floor. A force of \(40\,\text{N}\) acts on \(A\) towards the right, while a force of \(10\,\text{N}\) acts on \(C\) towards the left. What are the acceleration of the three-block system and the contact force exerted by \(B\) on \(C\)?
ⓐ. \(3\,\text{m s}^{-2}\) right; \(15\,\text{N}\) right
ⓑ. \(3\,\text{m s}^{-2}\) right; \(25\,\text{N}\) right
ⓒ. \(5\,\text{m s}^{-2}\) right; \(35\,\text{N}\) right
ⓓ. \(3\,\text{m s}^{-2}\) left; \(25\,\text{N}\) left
Correct Answer: \(3\,\text{m s}^{-2}\) right; \(25\,\text{N}\) right
Explanation: \( \textbf{Step 1 — Treat all three blocks as one system:} \) The contact forces between neighbouring blocks are internal to the combined system and therefore do not appear in its net external-force equation. Taking right as positive,
\[
F_{\text{ext}}=40-10=30\,\text{N}.
\]
The total mass is
\[
M=2+3+5=10\,\text{kg}.
\]
Hence,
\[
a=\frac{30}{10}=3\,\text{m s}^{-2}
\]
towards the right. Step 2 — Isolate block \(C\): Let the contact force exerted by \(B\) on \(C\) be \(F_{BC}\) towards the right. The external \(10\,\text{N}\) force acts leftward. Therefore,
\[
F_{BC}-10=m_Ca.
\]
Substituting,
\[
F_{BC}-10=5(3)=15.
\]
Thus,
\[
F_{BC}=25\,\text{N}.
\]
The \(40\,\text{N}\) applied force is not transmitted unchanged through every contact because different subsystems must accelerate different amounts of mass while also accounting for the opposing \(10\,\text{N}\) force. Hence the system acceleration is \(3\,\text{m s}^{-2}\) rightward and \(B\) exerts \(25\,\text{N}\) rightward on \(C\).
115. A \(50\,\text{kg}\) passenger stands on a scale in an elevator. During Stage I, the scale reads \(600\,\text{N}\). During Stage II, the scale reads \(400\,\text{N}\). Take \(g=10\,\text{m s}^{-2}\). Which statement is correct?
ⓐ. Stage I: \(2\,\text{m s}^{-2}\) upward; Stage II: \(2\,\text{m s}^{-2}\) downward; the direction of motion is undetermined.
ⓑ. Stage I: \(2\,\text{m s}^{-2}\) upward; Stage II: \(2\,\text{m s}^{-2}\) downward; the motion must reverse between stages.
ⓒ. Stage I: \(12\,\text{m s}^{-2}\) upward; Stage II: \(8\,\text{m s}^{-2}\) downward; the direction of motion is undetermined.
ⓓ. Stage I: zero acceleration; Stage II: zero acceleration; the direction of motion is undetermined.
Correct Answer: Stage I: \(2\,\text{m s}^{-2}\) upward; Stage II: \(2\,\text{m s}^{-2}\) downward; the direction of motion is undetermined.
Explanation: The passenger's weight is
\[
mg=50(10)=500\,\text{N}.
\]
The scale reading is the normal reaction \(N\), not the weight itself. Take upward as positive. In Stage I,
\[
N-mg=ma.
\]
Thus,
\[
600-500=50a,
\]
so
\[
a=2\,\text{m s}^{-2}
\]
upward. In Stage II,
\[
400-500=50a,
\]
giving
\[
a=-2\,\text{m s}^{-2}.
\]
Hence the acceleration is \(2\,\text{m s}^{-2}\) downward. The scale reading determines the acceleration direction because it reveals whether \(N\) is greater or smaller than \(mg\). It does not by itself reveal the direction of velocity. For example, an elevator moving downward can still have upward acceleration while slowing down. Therefore the accelerations can be inferred, but the directions of motion cannot.
116. A \(4\,\text{kg}\) block rests on a smooth plane inclined at \(30^\circ\) above the horizontal. A horizontal force \(P\) acts towards the right, and the plane rises towards the right. For one value of \(P\), the block has zero acceleration along the plane. Take \(g=10\,\text{m s}^{-2}\). What are \(P\) and the corresponding normal reaction \(N\)?
ⓐ. \(P=20\sqrt3\,\text{N}\), \(N=40\,\text{N}\)
ⓑ. \(P=\dfrac{40}{\sqrt3}\,\text{N}\), \(N=20\sqrt3\,\text{N}\)
ⓒ. \(P=\dfrac{40}{\sqrt3}\,\text{N}\), \(N=\dfrac{80}{\sqrt3}\,\text{N}\)
ⓓ. \(P=40\sqrt3\,\text{N}\), \(N=20\,\text{N}\)
Correct Answer: \(P=\dfrac{40}{\sqrt3}\,\text{N}\), \(N=\dfrac{80}{\sqrt3}\,\text{N}\)
Explanation: Choose axes parallel and perpendicular to the incline. The component of the weight down the plane is
\[
mg\sin30^\circ=4(10)\left(\frac12\right)=20\,\text{N}.
\]
The horizontal force has an up-slope component
\[
P\cos30^\circ=P\frac{\sqrt3}{2}.
\]
Zero acceleration along the plane requires
\[
P\frac{\sqrt3}{2}=20.
\]
Therefore,
\[
P=\frac{40}{\sqrt3}\,\text{N}.
\]
Now consider the direction perpendicular to the plane. The weight presses into the plane with component
\[
mg\cos30^\circ=40\frac{\sqrt3}{2}=20\sqrt3\,\text{N}.
\]
The horizontal force also has an into-plane component
\[
P\sin30^\circ=\frac{P}{2}
=\frac{20}{\sqrt3}\,\text{N}.
\]
Hence,
\[
N=20\sqrt3+\frac{20}{\sqrt3}.
\]
Writing the first term with common denominator,
\[
20\sqrt3=\frac{60}{\sqrt3},
\]
so
\[
N=\frac{80}{\sqrt3}\,\text{N}.
\]
Thus the force required to eliminate motion along the plane simultaneously increases the normal reaction above \(mg\cos30^\circ\).
117. Two bodies \(P\) and \(Q\) are tested separately. Their graphs of acceleration \(a\) against resultant force \(F\) are straight lines through the origin. The slope for \(P\) is \(0.50\,\text{kg}^{-1}\), while that for \(Q\) is \(0.20\,\text{kg}^{-1}\). The bodies are then connected and treated as one system. What resultant force is required to give the combined system an acceleration of \(4\,\text{m s}^{-2}\)?
ⓐ. \(12\,\text{N}\)
ⓑ. \(20\,\text{N}\)
ⓒ. \(28\,\text{N}\)
ⓓ. \(40\,\text{N}\)
Correct Answer: \(28\,\text{N}\)
Explanation: For a graph of acceleration against resultant force,
\[
a=\frac{F}{m}.
\]
Therefore the slope is
\[
\frac{a}{F}=\frac1m.
\]
For body \(P\),
\[
\frac1{m_P}=0.50\,\text{kg}^{-1},
\]
so
\[
m_P=2\,\text{kg}.
\]
For body \(Q\),
\[
\frac1{m_Q}=0.20\,\text{kg}^{-1},
\]
giving
\[
m_Q=5\,\text{kg}.
\]
The combined mass is therefore
\[
M=2+5=7\,\text{kg}.
\]
To produce acceleration \(4\,\text{m s}^{-2}\),
\[
F_{\text{net}}=Ma=7(4)=28\,\text{N}.
\]
The graph slopes must first be inverted to obtain the masses because the plotted graph is \(a\) versus \(F\), not \(F\) versus \(a\). Hence the required resultant is \(28\,\text{N}\).
118. Two blocks \(A\) and \(B\) of masses \(m\) and \(2m\) are connected by a light string on a smooth horizontal floor. A force \(F\) pulls \(B\) towards the right, and the system accelerates together. The string is then cut, while the force \(F\) continues unchanged on \(B\). Which statement correctly compares the accelerations immediately before and immediately after the cut?
ⓐ. Before the cut both accelerate at \(F/(3m)\); after the cut \(A\) has zero acceleration and \(B\) accelerates at \(F/(2m)\).
ⓑ. Before the cut \(A\) accelerates at \(F/m\) and \(B\) at \(F/(2m)\); after the cut both accelerate at \(F/(3m)\).
ⓒ. Before the cut both accelerate at \(F/(2m)\); after the cut \(A\) accelerates at \(F/m\) and \(B\) has zero acceleration.
ⓓ. Before and after the cut both blocks have acceleration \(F/(3m)\) because the external force is unchanged.
Correct Answer: Before the cut both accelerate at \(F/(3m)\); after the cut \(A\) has zero acceleration and \(B\) accelerates at \(F/(2m)\).
Explanation: Before the cut, the taut string constrains the two blocks to have a common acceleration. Treating both as one system, the only horizontal external force is \(F\), while the total mass is
\[
m+2m=3m.
\]
Thus,
\[
a_{\text{before}}=\frac{F}{3m}.
\]
The tension is internal to this combined system. When the string is cut, the interaction transmitted through the string disappears. Block \(A\) then has no horizontal force acting on it on the smooth floor, so
\[
a_A=0.
\]
Block \(B\) continues to experience the external force \(F\), but its own mass is only \(2m\). Therefore,
\[
a_B=\frac{F}{2m}.
\]
The velocity of either block does not change discontinuously at the cutting instant, but their accelerations do because their force sets change immediately. Hence option A is correct.
119. Three blocks \(A\), \(B\) and \(C\) of masses \(m\), \(2m\) and \(3m\) are connected in a line by light strings on a smooth horizontal surface. An external force \(F\) pulls \(C\) to the right. Let the tensions in the strings \(AB\) and \(BC\) be \(T_1\) and \(T_2\), respectively. What is the ratio \(T_1:T_2:F\)?
ⓐ. \(1:2:6\)
ⓑ. \(1:3:6\)
ⓒ. \(2:3:6\)
ⓓ. \(1:3:3\)
Correct Answer: \(1:3:6\)
Explanation: The total mass is
\[
m+2m+3m=6m.
\]
Therefore the common acceleration is
\[
a=\frac{F}{6m}.
\]
Now isolate block \(A\). The only horizontal force on it is \(T_1\), so
\[
T_1=ma=m\left(\frac{F}{6m}\right)=\frac{F}{6}.
\]
Next consider the subsystem \(A+B\), whose total mass is \(3m\). The only external horizontal force on this subsystem is \(T_2\), because \(T_1\) is internal when \(A+B\) is selected together. Hence,
\[
T_2=(3m)a.
\]
Substituting,
\[
T_2=3m\left(\frac{F}{6m}\right)=\frac{F}{2}.
\]
Therefore,
\[
T_1:T_2:F
=\frac{F}{6}:\frac{F}{2}:F
=1:3:6.
\]
The two string tensions are not equal because they belong to different strings and must accelerate different amounts of mass.
120. A light string passes over a smooth ideal pulley and connects masses \(4\,\text{kg}\) and \(2\,\text{kg}\). An additional downward force \(P\) is applied to the \(2\,\text{kg}\) mass. Take \(g=10\,\text{m s}^{-2}\). For what value of \(P\) does the system have zero acceleration, and what is the string tension then?
ⓐ. \(P=10\,\text{N}\), \(T=30\,\text{N}\)
ⓑ. \(P=20\,\text{N}\), \(T=40\,\text{N}\)
ⓒ. \(P=20\,\text{N}\), \(T=20\,\text{N}\)
ⓓ. \(P=40\,\text{N}\), \(T=40\,\text{N}\)
Correct Answer: \(P=20\,\text{N}\), \(T=40\,\text{N}\)
Explanation: If the acceleration is zero, each mass separately has zero resultant force. For the \(4\,\text{kg}\) mass,
\[
T-4g=0.
\]
Thus,
\[
T=4(10)=40\,\text{N}.
\]
For the \(2\,\text{kg}\) mass, the downward forces are its weight \(20\,\text{N}\) and the applied force \(P\), while the tension \(T=40\,\text{N}\) acts upward. Zero acceleration requires
\[
T-20-P=0.
\]
Therefore,
\[
40-20-P=0,
\]
so
\[
P=20\,\text{N}.
\]
The same result follows by treating both masses as one system: the \(40\,\text{N}\) weight on one side must balance the combined \(20\,\text{N}+P\) downward driving influence on the other side. The tension is not obtained by adding the two weights; it follows from the force balance on either individual mass. Hence \(P=20\,\text{N}\) and \(T=40\,\text{N}\).
