Question

A ball (A) of mass 1.4kg is moving with a speed of 3.7m/s collides with a
0.55kg ball (B) which is at rest. If Ball A is going 2.3m/s after the collision,
what will the velocity of Ball B be after the collision?​

Answer 1

Answer:

3.57 m/s

Explanation:

Total momentum remains conserved.

Before that collision*,

Velocity of A and B is 3.7 and 0 respectively.

Total momentum = (m1v1) + (m2v2)

= (1.4 x 3.7) + (0.55 x 0) = 5.18

After that collision,

Velocity of A is 2.3 & velocity of B be 'v',

Total momentum = (m1 x 2.3) + (m2 v)

= (1.4 x 2.3) + (0.55v) = 3.22 + 0.55v

Since momentum remains conserved,

Initial momentum = final momentum

5.18 = 3.22 + 0.55v

1.96 = 0.55v

3.57 = v

Velocity of the ball B is 3.57 m/s

Answer 2

GiveN :-

• A ball A of mass 1.4kg is moving with a speed of 3.7m/s collides with a 0.55kg ball B which is at rest.
• Ball A is going 2.3m/s after the collision .

To FinD :-

• The velocity of Ball B be after the collision.

SolutioN :-

Given that , A ball A of mass 1.4kg is moving with a speed of 3.7m/s collides with a 0.55kg ball B which is at rest. The velocity of Ball B be after the collision.We need to find the velocity of Ball B be after the collision. Here we can use law of conservation of momentum according to which total momentum in a system remains conserved . Let the velocity of the Ball B after collision be v.

★ Using Law of Conservation of Momentum:-

$\sf:\dashrightarrow \pink{ m_1u_1+m_2u_2=m_1v_1+m_2v_2 }\\\\\sf\dashrightarrow (1.4kg)(3.7\ m/s) + (0.55 \ kg)(0m/s)= (1.4 kg)(2.3)+ v(0.55 kg) \\\\\sf:\dashrightarrow 5.18 kg - m/s + 0 = 3.22 kg-m/s + (0.55 kg)v \\\\\sf:\dashrightarrow 5.18 \ kg-m/s - 3.22 \ kg-m/s = v (0.55kg) \\\\\sf:\dashrightarrow v(0.55 kg) = 1.96 kg- m/s\\\\\sf:\dashrightarrow v =\dfrac{1.96 \ kg-m/s}{0.55kg} \\\\\sf:\dashrightarrow\underset{\blue{\sf Required\ Velocity }}{\underbrace{\boxed{\pink{\frak{ Velocity_{(Ball \ B )}= 3.563 \ m/s }}}}}$