Collision and Impulse MCQ Quiz - Objective Question with Answer for Collision and Impulse - Download Free PDF

Correct option is: (1) 21 NS

v₁ = √(2gh₁)

= √(2 × 9.8 × 40)

= √784 = 28 m/s

v₂ = √(2gh₂) = √(2 × 9.8 × 10)

= √196 = 14 m/s

Impulse = Δp = m(v_f − v_i) = m(v₂ − (−v₁))

= (1/2)(14 − (−28))

Concept:

Impulse

• When a large force works on a body for a very small time interval, it is called impulsive force.
• An impulsive force does not remain constant, but changes first from zero to maximum and then from maximum to zero. In such case, we measure the total effect of force.
• The impulse caused by a force during a specific time interval is equal to the body's change of momentum during that time interval.
• Impulse, effectively, is a measure of change in momentum.
• Impulse of a force is a measure of the total effect of force.
• Impulse is a vector quantity and its direction is same as that of force.
• SI unit of impulse is Newton-second or kg-m-s-1.

$\overrightarrow{I}={\int }_{t_1}^{t_2}\overrightarrow{F}\cdot dt$

Where, I = impulse, F = force and dt = very small time interval

Hence according to given explanation above

• Impulse, $I=\int \frac{dp}{dt}.dt\Rightarrow i.e.,I=(p_2-p_1)=\Delta p$
• therefore we can say that impulse is equals to change in momentum

Calculation:

Mass of the ball (m) = 150 g = 0.15 kg

Initial velocity (u) = 12 m/s

Final velocity (v) = 20 m/s

Time = 0.01 s

Initial Momentum p₁ = 0.15 × (-12) = -1.8 kg m/s

Final Momentum p₂ = 0.15 kg × 20 = 3 kg m/s

Impulse = F × T = change of momentum

⇒ F × T = p₂ - p₁ = (3 - (-1.8)) = 4.8 

⇒ F × 0.01 = 4.8

⇒ F = 480 N -s

Calculation:

Given that a bullet of mass 'a' and velocity 'b' is fired into a large block of wood of mass 'c'. The bullet gets embedded into the block of wood. To find the final velocity of the system, we will use the principle of conservation of momentum.

According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision, assuming no external forces act on the system.

The initial momentum of the system is:

Momentum before = mass of bullet × velocity of bullet = a × b

The final momentum of the system after the bullet gets embedded in the block is:

Momentum after = (mass of bullet + mass of block) × final velocity = (a + c) × v

By conservation of momentum, we equate the initial momentum to the final momentum:

a × b = (a + c) × v

Solving for the final velocity v:

v = (a × b) / (a + c)

Concept:

In a perfectly elastic collision, momentum and kinetic energy are conserved.

The velocities after the collision along the line of impact (defined by $\hat{n}$) are updated according to the following formula:

${\overrightarrow{v}}_A^{\prime }={\overrightarrow{v}}_A-\frac{2m_B}{m_A+m_B}[({\overrightarrow{v}}_A-{\overrightarrow{v}}_B)\cdot \hat{n}]\hat{n}$

To find the final speed of A, calculate its velocity vector ${\overrightarrow{v}}_A^{\prime }$ after the collision and take the magnitude.

Calculation:

Step 1: Normalize the line of impact vector:

$\overrightarrow{n}=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}$

This vector is already normalized, so $\hat{n}=\overrightarrow{n}$.

Step 2: Compute the relative velocity along the line of impact:

$$\begin{gathered} {\overrightarrow{v}}_A=(3\hat{i}-2\hat{j}) \\ {\overrightarrow{v}}_B=0 \end{gathered}$$

${\overrightarrow{v}}_A-{\overrightarrow{v}}_B=(3\hat{i}-2\hat{j})$

$({\overrightarrow{v}}_A-{\overrightarrow{v}}_B)\cdot \hat{n}=(3\hat{i}-2\hat{j})\cdot (\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j})$

$=\frac{3}{\sqrt{2}}-\frac{2}{\sqrt{2}}=\frac{1}{\sqrt{2}}$

Step 3: Compute the final velocity of A:

${\overrightarrow{v}}_A^{\prime }={\overrightarrow{v}}_A-\frac{2m_B}{m_A+m_B}\cdot \left(\frac{1}{\sqrt{2}}\right)\hat{n}$

${\overrightarrow{v}}_A^{\prime }=(3\hat{i}-2\hat{j})-\frac{2(2)}{4+2}\cdot \frac{1}{\sqrt{2}}\cdot (\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j})$

${\overrightarrow{v}}_A^{\prime }=(3\hat{i}-2\hat{j})-\frac{4}{6}\cdot (\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j})$

${\overrightarrow{v}}_A^{\prime }=(3\hat{i}-2\hat{j})-\frac{2}{3}\cdot (\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j})$

${\overrightarrow{v}}_A^{\prime }=(3-\frac{2}{3\sqrt{2}})\hat{i}+(-2-\frac{2}{3\sqrt{2}})\hat{j}$

Step 4: Compute the magnitude of the final velocity of A:

$|{\overrightarrow{v}}_A^{\prime }|=\sqrt{{(3-\frac{2}{3\sqrt{2}})}^2+{(-2-\frac{2}{3\sqrt{2}})}^2}$

Simplify to find:

$|{\overrightarrow{v}}_A^{\prime }|\approx 3.3\text{m/s}$

∴ The final speed of A after the collision is approximately 3.3 m/s

Elastic Collision Between Two Particles (Similar Concept)

Given:

• Mass of particle A=m

• Mass of particle B=3m

• Initial velocity of A=$v\hat{i}$

• Particle B is initially at rest

• After collision, A moves with velocity $v_A\hat{j}$

Solution:

Using the principle of conservation of momentum and the fact that the collision is elastic (both kinetic energy and momentum are conserved):

1. Conservation of Momentum:

In the x-direction:

$mv=3m{v}_B\cos \theta$

In the y-direction (since particle B was initially at rest and moves only after the collision):

$0=m{v}_A-3m{v}_B\sin \theta$

2. Conservation of Kinetic Energy:

$\frac{1}{2}m{v}^2=\frac{1}{2}m{v}_A^2+\frac{1}{2}3m{v}_B^2$

This simplifies to:

$v^2=v_A^2+3v_B^2$

After solving, (1), (2) and (3) we get:

$v_A^2=3v_B^2$

The answer is 3.

CONCEPT:

• Impulse: When a large force works on a body for a very small time interval, it is called impulsive force.
  • The impulse caused by a force during a specific time interval is equal to the body's change of momentum during that time interval.
  • Impulse, effectively, is a measure of the change in momentum. It is denoted by J or I.

J or I = Δp = F × Δt = Δ (m v) = m (Δ v)

Where Δp = change in momentum, F =force, and Δt = change in time

• The SI unit of Impulse is Ns.
• The dimensional formula pf Impulse is [MLT-1].

EXPLANATION:

Since 

I = F × Δt (Δt = Final time - Initial time)

I = m × Δv (Δv = Final velocity - Initial velocity)

So option 3 is correct.

Concept:

Impulse:

If a Force F is applied for a small time interval t, then the product of F and t is known as Impulse.

Numerically, Impulse is equal to change in momentum.

F × t = m (v – u)

where v = velocity after the impact, u = velocity before the impact, t = fraction of time during which force is applied

Calculation:

Given:

F = 50 N, t = 10 ms = 10 × 10^-3 s, u = 0 and m = 0.5 kg

F × t = m (v – u)

50 × 10 × 10^-3 = 0.5 × (v – 0)

CONCEPT:

• Momentum: Product of the mass of a particle and its velocity.
  • Momentum is a vector quantity: I .e.., it has both magnitude and direction
  • Isaac Newton's second law of motion states that the time rate of change of momentum is equal to the force acting on the particle.
• ​Impulse: change in momentum equals the average net external force multiplied by the time this force acts.

ΔP = F_net  Δt

EXPLANATION:

Let initially the body is moving along the positive x-axis with speed v.

so, the momentum of the body initially

P_i = mv 

• After bouncing back, the body moves along the negative x-axis at the same speed.
• so, the momentum of the body finally

P_f =  -mv

• Thus change in momentum of the body:

 ΔP = P_f - P_i = -2mv

• Impulse experienced 

ΔP = -2mv

• The magnitude of impulse experienced = 2mv.

​option 4 is the answer.

Concept:

Impulse

• When a large force works on a body for a very small time interval, it is called impulsive force.
• An impulsive force does not remain constant, but changes first from zero to maximum and then from maximum to zero. In such case, we measure the total effect of force.
• The impulse caused by a force during a specific time interval is equal to the body's change of momentum during that time interval.
• Impulse, effectively, is a measure of change in momentum.
• Impulse of a force is a measure of the total effect of force.
• Impulse is a vector quantity and its direction is same as that of force.
• SI unit of impulse is Newton-second or kg-m-s-1.

$\overrightarrow{I}={\int }_{t_1}^{t_2}\overrightarrow{F}\cdot dt$

Where, I = impulse, F = force and dt = very small time interval

Hence according to given explanation above

• Impulse, $I=\int \frac{dp}{dt}.dt\Rightarrow i.e.,I=(p_2-p_1)=\Delta p$
• therefore we can say that impulse is equals to change in momentum

Calculation:

Mass of the ball (m) = 150 g = 0.15 kg

Initial velocity (u) = 12 m/s

Final velocity (v) = 20 m/s

Time = 0.01 s

Initial Momentum p₁ = 0.15 × (-12) = -1.8 kg m/s

Final Momentum p₂ = 0.15 kg × 20 = 3 kg m/s

Impulse = F × T = change of momentum

⇒ F × T = p₂ - p₁ = (3 - (-1.8)) = 4.8 

⇒ F × 0.01 = 4.8

⇒ F = 480 N -s

CONCEPT:

• Perfectly elastic collision: A perfectly elastic collision is defined as one in which there is no loss of kinetic energy and momentum of the system is conserved  in the collision.
• Inelastic collision: A inelastic collision is defined as one in which there is a loss of kinetic energy and momentum of the system is conserved in the collision.
• Coefficient of restitution is the ratio of relative velocity after impact to the relative velocity before impact. 

Coefficient of restitution (e)

$\mathrm{e}=\frac{\mathrm{R}\mathrm{e}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{l}\mathrm{o}\mathrm{c}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{a}\mathrm{f}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{c}\mathrm{o}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{s}\mathrm{i}\mathrm{o}\mathrm{n}}{\mathrm{R}\mathrm{e}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{v}\mathrm{e}\mathrm{l}\mathrm{o}\mathrm{c}\mathrm{i}\mathrm{t}\mathrm{y}\mathrm{b}\mathrm{e}\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{s}\mathrm{i}\mathrm{o}\mathrm{n}}=\frac{{\mathrm{v}}_2-{\mathrm{v}}_1}{{\mathrm{u}}_1-{\mathrm{u}}_2}$

• For perfectly elastic collision, e = 1
• For inelastic collision, e < 1
• For a perfectly inelastic collision,e = 0

​EXPLANATION:

As from the above discussion, we can say

• The coefficient of restitution for perfectly inelastic bodies is: e = 0

​So option 1 is correct.

CONCEPT:

Impulse:

• When a large force works on a body for a very small time interval, it is called impulsive force.
• An impulsive force does not remain constant, but changes first from zero to maximum and then from maximum to zero. In such a case we measure the total effect of force.
• The impulse caused by a force during a specific time interval is equal to the body's change of momentum during that time interval.
• Impulse, effectively, is a measure of the change in momentum.
• The impulse of a force is a measure of the total effect of force.
  $\overrightarrow{I}={\int }_{t_1}^{t_2}\overrightarrow{F}\cdot dt$
  Where, I = impulse, F = force and dt = very small time interval
• Impulse is a vector quantity and its direction is the same as that of force.
• SI unit of impulse is Newton-second or kg-m-s-1.

EXPLANATION:

• The linear momentum of a body is the quantity of motion contained in the body and it is measured as the product of the mass of the body and its velocity. Therefore 1 is incorrect.
• From above it is clear that an impulse is equivalent to a force applied to an object for a short period of time. Thus option 2 is correct.
• Tension is a force that acts along the lengths of a rope or cable. Therefore option 3 is incorrect.

CONCEPT:

Linear Momentum:

• The linear momentum of a body is the quantity of motion contained in the body.
• It is measured in terms of the force required to stop the body in unit time.
• It is also measured as the product of the mass of the body and its velocity i.e.,

⇒ Momentum (P) = mass (m) × velocity (v)

⇒ P = mv

Impulse:

• When a large force is acting on a body for a short period of time to produce a finite change in its momentum is called impulsive force.
• The impulse is given as,

⇒ I = F×dt

• Impulse is also equal to the change in momentum.

EXPLANATION:

Given m = 10 kg, v₁ = 20 m/sec, v2 = 30 m/sec, and dt = 0.6 sec

• The momentum is given as,

⇒ P = mv     -----(1)

• So initial momentum is given as,

⇒ P₁ = mv₁

⇒ P1 = 10 × 20 

⇒ P1 = 200 kg-m/sec     -----(2)

• And final momentum is given as,

⇒ P2 = mv2

⇒ P2 = 10 × 30

⇒ P2 = 300 kg-m/sec     -----(3)

By equation 2 and equation 3 the change in momentum is given as,

⇒ ΔP = P₂ - P₁

⇒ ΔP = 300 - 200

⇒ ΔP = 100 kg-m/sec

• We know that the impulse is equal to the change in momentum,

⇒ I = ΔP

⇒ I = 100 kg-m/sec

• Hence, option 2 is correct.

CONCEPT:

Impulse: When a large force works on a body for a very small time interval, it is called impulsive force.

• The impulse caused by a force during a specific time interval is equal to the body's change of momentum during that time interval.
• Impulse, effectively, is a measure of the change in momentum. It is denoted by J or I.

J or I = Δp = F × Δt = Δ (mv) = m (Δv)

where Δp = change in momentum, F =force, and Δt = change in time

• The SI unit of Impulse is Ns.
• The dimensional formula of Impulse is [MLT^-1].

EXPLANATION:

Given that:

Impulse (J) = change in momentum (Δp) = 1.8 kg m/s

Time (Δt) = 0.2 sec

Since J = F × Δ t

Force (F) = J/Δ t = 1.8/0.2 = 9 N

Hence option 1 is correct.

CONCEPT:

• Elasticity: The property of a body by virtue of which the body can recover its original shape and size after removal of the deforming force.

• Momentum (p): The product of mass and velocity is called as the momentum of the body.

Momentum (p) = m × V

Where, m = mass of the body, and V = velocity of the body.

CALCULATION:

Given that,

For rubber ball:

The initial momentum of the rubber ball as it was thrown towards the wall, p1 = p

Final momentum of the rubber ball as it bounces back from the wall, p2 = -p

Whereas the net change in momentum Δp can be given as

$\mathrm{\Delta }p=p_1-p_2=p-\left(-p\right)=2p$

For metal ball:

The initial momentum of the rubber ball as it was thrown towards the wall, p1 = p

Final momentum of the rubber ball as it bounces back from the wall, p2 = 0 (Since it comes in rest after striking the wall)

Whereas the net change in momentum Δp can be given as

$\mathrm{\Delta }p=p_1-p_2=p-\left(0\right)=p$

• Since the rubber is more elastic than the metal. that's why the rubber ball rebounds and the metal ball comes in rest. So the assertion is correct.
• As the change in momentum of the rubber ball is more than that in the case of metal ball. So the reason is also correct. But the reason is not the correct explanation for the assertion because the correct reason is elasticity. Hence option 2 is correct.

CONCEPT:

Impulse:

• When a large force works on a body for a very small time interval, it is called impulsive force.
• The impulse caused by a force during a specific time interval is equal to the body's change of momentum during that time interval.
• Impulse, effectively, is a measure of the change in momentum.

Δp = F × Δt

Where Δp = change in momentum, F =force, and Δt = change in time

EXPLANATION:

• Impulse is a vector quantity and its direction is the same as that of force.
• SI unit of impulse is Newton-second or kg-m-s-1.

CALCULATION:

Given - mass (m) = 4 kg, initial velocity (v₁) = final velocity (v₂) = 5 m/s  and time (Δt) = 0.05 s

• Initial momentum of the ball is 

⇒ p₁ = mv₁ = 4 × 5 = 20 kgm/s     ------ (1)

• Final momentum of the ball is 

⇒ p2 = mv2 = 4 × (-5) = -20 kgm/s     ------ (2)        [∵ because it rebounds elastically]

• The change in the momentum is

⇒ Δp = p1 - (p2) = 20 + 20 = 40 kgm/s

• Impulse, effectively, is a measure of the change in momentum.

⇒ Δp = F × Δt

$\Rightarrow F=\frac{\Delta p}{\Delta t}=\frac{40}{0.05}=800N$