30 Momentum, Impulse and Collisions MCQs with Answers

Linear momentum is defined as

\[
\vec{p}=m\vec{v}
\]

Mass has unit \(\text{kg}\), while velocity has unit \(\text{m/s}\).

Therefore, momentum has unit

\[
\text{kg}\cdot\text{m/s}
\]

Momentum is a vector quantity, so direction is also important.

Use

\[
\vec{p}=m\vec{v}
\]

\[
p=2\times5=10\text{ kg}\cdot\text{m/s}
\]

Momentum has the same direction as velocity.

Therefore, the momentum is east.

Impulse is

\[
J=F\Delta t
\]

\[
J=12\times0.50
\]

\[
J=6\text{ N}\cdot\text{s}
\]

Impulse equals the change in momentum.

Take right as positive.

Initial momentum:

\[
p_i=0.20(10)=2\text{ kg}\cdot\text{m/s}
\]

Final momentum:

\[
p_f=0.20(-5)=-1\text{ kg}\cdot\text{m/s}
\]

Change in momentum:

\[
\Delta p=p_f-p_i
\]

\[
\Delta p=-1-2=-3\text{ kg}\cdot\text{m/s}
\]

The magnitude is

\[
|\Delta p|=3\text{ kg}\cdot\text{m/s}
\]

For a system,

\[
\Delta\vec{p}_{\text{system}}=\vec{J}_{\text{external}}
\]

If the net external impulse is zero, then

\[
\Delta\vec{p}_{\text{system}}=0
\]

Therefore, total momentum remains constant.

Internal forces can change individual momenta, but they cannot change total system momentum.

Initially, total momentum is zero.

After firing,

\[
m_bv_b+m_rv_r=0
\]

\[
0.020(400)+4v_r=0
\]

\[
8+4v_r=0
\]

\[
v_r=-2.0\text{ m/s}
\]

The negative sign shows that the rifle moves opposite to the bullet.

Therefore, recoil speed is

\[
2.0\text{ m/s}
\]

Take the direction of the 2 kg cart as positive.

\[
m_1=2,\qquad v_1=4
\]

\[
m_2=3,\qquad v_2=-2
\]

Since the carts stick together,

\[
m_1v_1+m_2v_2=(m_1+m_2)v_f
\]

\[
2(4)+3(-2)=5v_f
\]

\[
8-6=5v_f
\]

\[
v_f=0.4\text{ m/s}
\]

The positive sign shows that the combined carts move in the direction of the 2 kg cart.

In every isolated collision,

\[
\text{Total momentum is conserved}
\]

However, kinetic energy is conserved only in a perfectly elastic collision.

Therefore, for a perfectly elastic collision:

\[
p_i=p_f
\]

and

\[
K_i=K_f
\]

Initial momentum:

\[
p_i=1(6)=6
\]

Final velocity:

\[
v_f=\frac{6}{1+2}=2\text{ m/s}
\]

Initial kinetic energy:

\[
K_i=\frac{1}{2}(1)(6^2)=18\text{ J}
\]

Final kinetic energy:

\[
K_f=\frac{1}{2}(3)(2^2)=6\text{ J}
\]

Energy lost:

\[
18-6=12\text{ J}
\]

Fraction lost:

\[
\frac{12}{18}=\frac{2}{3}
\]

Impulse equals the area under the force-time graph.

The graph is a triangle.

\[
J=\frac{1}{2}\times\text{base}\times\text{height}
\]

\[
J=\frac{1}{2}\times4\times20
\]

\[
J=40\text{ N}\cdot\text{s}
\]

The bullet sticks to the block, so this is a perfectly inelastic collision.

\[
m_bv_b=(m_b+m_B)v_f
\]

\[
0.010(500)=(0.010+1.99)v_f
\]

\[
5=2v_f
\]

\[
v_f=2.5\text{ m/s}
\]

Momentum is conserved during the short collision.

For a one-dimensional elastic collision with the second cart initially at rest,

\[
v_1=
\frac{m_1-m_2}{m_1+m_2}u_1
\]

\[
v_1=
\frac{2-1}{2+1}(6)
\]

\[
v_1=2\text{ m/s}
\]

The heavier cart continues forward, but with lower speed.

For a one-dimensional elastic collision where the target is initially at rest,

\[
v_2=
\frac{2m_1}{m_1+m_2}u_1
\]

\[
v_2=
\frac{2(1)}{1+3}(8)
\]

\[
v_2=4\text{ m/s}
\]

The lighter object transfers part of its momentum and kinetic energy to the heavier object.

Take right as positive.

\[
p_i=4(3)+2(-6)
\]

\[
p_i=12-12=0
\]

Since the carts stick together,

\[
(m_1+m_2)v_f=0
\]

\[
v_f=0
\]

The system remains at rest after collision because total initial momentum is zero.

Take upward as positive.

Initial velocity:

\[
v_i=-10\text{ m/s}
\]

Final velocity:

\[
v_f=8\text{ m/s}
\]

Impulse equals change in momentum:

\[
J=m(v_f-v_i)
\]

\[
J=0.20[8-(-10)]
\]

\[
J=0.20(18)
\]

\[
J=3.6\text{ N}\cdot\text{s}
\]

Momentum is conserved.

Initial momentum:

\[
p_i=0.020(300)=6
\]

Final bullet momentum:

\[
p_b=0.020(100)=2
\]

Thus, block momentum is

\[
p_B=6-2=4
\]

Since block mass is 1 kg,

\[
v_B=\frac{4}{1}=4\text{ m/s}
\]

After collision, the combined mass rises by \(h=0.20\text{ m}\).

Use energy after collision:

\[
\frac{1}{2}(M+m)V^2=(M+m)gh
\]

\[
V=\sqrt{2gh}
\]

\[
V=\sqrt{2(10)(0.20)}=2\text{ m/s}
\]

Now use momentum during collision:

\[
mu=(M+m)V
\]

\[
0.020u=(2.020)(2)
\]

\[
u=202\text{ m/s}
\]

Therefore, the correct bullet speed is approximately

\[
202\text{ m/s}
\]

The options need correction before publishing. The correct option should be close to \(200\text{ m/s}\).

The shell starts from rest, so total momentum is zero.

Take east as positive.

\[
2(12)+3v=0
\]

\[
24+3v=0
\]

\[
v=-8\text{ m/s}
\]

The negative sign means west.

Initial momentum components are

\[
p_x=2(3)=6
\]

\[
p_y=4(2)=8
\]

Total mass is

\[
M=2+4=6
\]

Final velocity components:

\[
v_x=\frac{6}{6}=1
\]

\[
v_y=\frac{8}{6}=\frac{4}{3}
\]

Therefore,

\[
v=\sqrt{1^2+\left(\frac{4}{3}\right)^2}
\]

\[
v=\frac{5}{3}\text{ m/s}
\]

Take the initial direction as positive.

\[
p_i=mu
\]

After rebound,

\[
p_f=-mu
\]

Change in momentum:

\[
\Delta p=p_f-p_i
\]

\[
\Delta p=-mu-mu=-2mu
\]

The magnitude of impulse is

\[
|\Delta p|=2mu
\]

For a one-dimensional elastic collision with the second particle initially at rest,

\[
v_1=\frac{m_1-m_2}{m_1+m_2}u
\]

Here,

\[
m_1=m,\qquad m_2=3m
\]

Therefore,

\[
v_1=\frac{m-3m}{m+3m}u
\]

\[
v_1=-\frac{u}{2}
\]

The first particle rebounds with speed

\[
\frac{u}{2}
\]

Its final kinetic energy is

\[
K_f=\frac{1}{2}m\left(\frac{u}{2}\right)^2
\]

\[
K_f=\frac{1}{4}\left(\frac{1}{2}mu^2\right)
\]

Hence, the fraction of initial kinetic energy remaining with the first particle is

\[
\frac{1}{4}
\]

During the collision, momentum is conserved.

\[
mu=(M+m)V
\]

\[
V=\frac{mu}{M+m}
\]

After collision, the combined mass is

\[
M+m
\]

Its angular frequency is

\[
\omega=\sqrt{\frac{k}{M+m}}
\]

The collision occurs at equilibrium, so the speed is maximum.

Therefore,

\[
A=\frac{V}{\omega}
\]

\[
A=
\frac{\frac{mu}{M+m}}
{\sqrt{\frac{k}{M+m}}}
\]

\[
A=
\frac{mu}{\sqrt{k(M+m)}}
\]

So the correct result is

\[
\frac{mu}{\sqrt{k(M+m)}}
\]

For an elastic collision between identical masses, the objects exchange velocities.

Initial velocities are

\[
v_1=u
\]

and

\[
v_2=-2u
\]

Therefore, after collision,

\[
v_1=-2u
\]

\[
v_2=u
\]

At the highest point, the shell has horizontal velocity

\[
u\cos\theta
\]

Before explosion, momentum is

\[
Mu\cos\theta
\]

One fragment stops. The other fragment has mass

\[
\frac{M}{2}
\]

Using horizontal momentum conservation,

\[
Mu\cos\theta=
\frac{M}{2}v
\]

\[
v=2u\cos\theta
\]

The time from the highest point to the ground is

\[
t=\frac{u\sin\theta}{g}
\]

The second fragment travels horizontally after explosion:

\[
x_2=
2u\cos\theta
\cdot
\frac{u\sin\theta}{g}
\]

\[
x_2=
\frac{u^2\sin2\theta}{g}
\]

The highest point lies at horizontal distance

\[
x_h=
\frac{u^2\sin2\theta}{2g}
\]

Therefore, total range of the second fragment is

\[
x_h+x_2
=
\frac{3u^2\sin2\theta}{2g}
\]

Take the initial direction as positive.

Initial momentum:

\[
p_i=mu
\]

After collision, the particle rebounds, so

\[
p_f=-meu
\]

Change in momentum:

\[
\Delta p=p_f-p_i
\]

\[
\Delta p=-meu-mu
\]

\[
\Delta p=-m(1+e)u
\]

Therefore, impulse magnitude is

\[
m(1+e)u
\]

Speed just before impact is

\[
v=\sqrt{2gh}
\]

After rebound, speed becomes

\[
v’=ev
\]

The rebound height is

\[
H=\frac{v’^2}{2g}
\]

\[
H=\frac{e^2v^2}{2g}
\]

Since

\[
v^2=2gh
\]

\[
H=e^2h
\]

For the first block to stop after an elastic collision,

\[
v_1=0
\]

Using the elastic collision formula,

\[
v_1=
\frac{m-M}{m+M}u
\]

For

\[
v_1=0
\]

we need

\[
m-M=0
\]

Therefore,

\[
M=m
\]

This is why identical balls exchange velocities in a one-dimensional elastic collision.

The rocket pushes fuel backward, while the fuel pushes the rocket forward.

These are internal forces within the rocket-plus-fuel system.

If external forces are ignored,

\[
\Delta p_{\text{system}}=0
\]

Therefore, total momentum remains constant.

The rocket alone gains forward momentum, while expelled fuel gains equal backward momentum.

Initial momentum:

\[
\vec p_i=mu\hat{i}
\]

Each fragment has mass

\[
\frac{m}{2}
\]

Let velocity of second fragment be \(\vec v\).

Momentum conservation gives

\[
mu\hat{i}
=
\frac{m}{2}(u\hat{j})
+
\frac{m}{2}\vec v
\]

Multiply by

\[
\frac{2}{m}
\]

\[
2u\hat{i}
=
u\hat{j}
+
\vec v
\]

Therefore,

\[
\vec v=
2u\hat{i}-u\hat{j}
\]

Take the mass \(m\) moving right and \(2m\) moving left.

Initial momentum:

\[
p_i=mu-2mu=-mu
\]

Combined mass:

\[
3m
\]

Final velocity:

\[
v_f=
\frac{-mu}{3m}
=
-\frac{u}{3}
\]

Initial kinetic energy:

\[
K_i=
\frac{1}{2}mu^2
+
\frac{1}{2}(2m)u^2
\]

\[
K_i=
\frac{3}{2}mu^2
\]

Final kinetic energy:

\[
K_f=
\frac{1}{2}(3m)
\left(\frac{u}{3}\right)^2
\]

\[
K_f=
\frac{1}{6}mu^2
\]

Fraction remaining:

\[
\frac{K_f}{K_i}
=
\frac{1/6}{3/2}
=
\frac{1}{9}
\]

Therefore, fraction lost is

\[
1-\frac{1}{9}
=
\frac{8}{9}
\]