Physics, asked by alondracruz5771, 5 hours ago

A 0.80-kg ball is thrown with a velocity of 13 m/s. The ball hit a 0.12-kg bowling pin. When the ball hits the pin, the pin goes flying forward at 6 m/s. What is the velocity of the ball after it hits the pin?

Answers

Answered by parvatijawargi458
0

Answer:

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Answered by GulabLachman
1

Given: A 0.80 kg ball is thrown with a velocity of 13 m/s. The ball hits 0.12 kg bowling pin. When the ball hits the pin, it goes flying forward at 6 m/s

To find: Velocity of ball after it hits the pin

Explanation: Mass of ball (m1) = 0.80 kg

Mass of pin(m2) = 0.12 kg

Initial velocity of the ball(u1) = 13 m/s

Initial velocity of the pin(u2) = 0 ( at rest)

Final velocity of the pin (v2)= 6 m/s

Let final velocity of ball be v.

Using conservation of momentum which is valid for every type of collision:

m1u1+ m2u2= m1v + m2v2

=> 0.8*13 + 0.12*0 = 0.8* v + 0.12* 6

=> 10.4 +0 = 0.8 v + 0.72

=> 10.4-0.72 = 0.8 v

=> v = 9.68/0.8

= 12.1 m/s

Therefore, the final velocity of the ball is 12.1 m/s after it hits the pin.

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