Two masses mA = 6 kg and mB = 4 kg collide. Before the collision, the mass A goes to the right at 5 m/s while B goes to the left at 5 m/s. Left at 12 m/s. After the collision, mass B rebounds with a speed of 1 m/s to the right.

a. What is the speed of A after the collision?

b. Was the collision elastic?

c. If the collision lasted 0.1 s, what is the force of impact (assumed to be 0.1 s) on A?

constant)?

According to the question,

 

Given: mass A (m_A) = 6kg

mass B (m_B) = 4 kg

 

Velocity of mass A before collision (V_A) = +5m/s

 

Velocity of mass B before collision (V_B) = -12 m/s

 

Note: Here we are considering right direction as positive (+) and left direction as negative (-).

 

Velocity of mass B after collision (V_B') = +1 m/s

 

(a) Let us consider velocity of mass A after collision = V_A'

 

Applying conservation of momentum principle-

 

Momentum Before collision = Momentum After collision

 

m_AV_A + m_BV_B = m_AV_A' + m_BV_B'

 

6 * (+5) + 4 * (-12) = 6 * V_A' + 4 * (+1)

 

30 - 48 = 6 * V_A' + 4

 

6 V_A' = -22

 

V_A' = -22 / 6

 

V_A' = -3.67 m/s

 

Therefore, speed of A after collision is 3.67 m/s.

Direction of A is towards the left.

 

 

(b) To know whether the collision is elastic or inelastic, we will find the Kinetic energy of the system (A + B) before and after collision.

 

If Kinetic energy remains conserved, the collision will be elastic.

 

If Kinetic energy does not remain conserved i.e., there is a Range in Kinetic energy before and after collision, the collision will be elastic.

 

Kinetic energy of the system before collision-

 

K_i = 1/2 m_AV_A² + 1/2 m_BV_B²

 

K_i = 1/2 * 6 * (5)² + 1/2 * 4 * (-12)²

 

K_i = 75 + 288

 

K_i = 363 Joule ----- (1)

 

Final Kinetic energy of the system after collision

 

K_f = 1/2 m_AV_A'² + 1/2 m_BV_B'²

 

= 1/2 * 6 (-3.67)² + 1/2 * 4 * (1)²

 

= 40.41 + 2

 

K_f = 42.41 Joule ------ (2)

 

From equation (1) and equation (2) we can see that there is a change in Kinetic energy of the system, before and after collision.

 

This means Kinetic energy is not considered.

 

Thus, we can conclude that the collision is not elastic.

 

 

(c) Collision time (Δt) = 0.1 sec

 

Let the Force of impact is = F

 

Using Impulse - Momentum Principle -

 

F.Δt = Final Momentum - Initial Momentum

 

F.Δt =(m_AV_A' + m_BV_B') - (m_AV_A + m_BV_B)

 

F.(0.1) = [6 * (*3.67) + 4 * 1] - [6 * (+5) + 4 * (-12)]

 

F.(0.1) = (-22.02 + 4) - (30 - 48)

 

F.(0.1) = (-18.02) - (-18)

 

F.(0.1) = -18.02 + 18

 

F = -0.02 / 0.1

 

F = 0.2 Newton (Approximately)

 

Force of Impact is 0.2 Newton (Approximately)