Question

PLEASE ANSWER ASAP

A 5 kg bowling ball traveling at 2 m/s hits a motionless 10 kg bowling ball. If the smaller ball bounces back at a speed of -1 m/s, what will be the speed of the larger ball after the collision?

Answer 1

Given ↷

• m1 =5kg
• m2=10( since it's at rest)
• v1=2m/s
• v2=0(since it's at rest)
• v1' = -1

To find ↷

• the speed of the larger ball after the collision

Solution↷

We know that,

m1v1 + m2v2 = m1v1' + m2v2'

5 x 2 + 0 = 5 x -1 + 10 x v2'

10= -5 +10 x v2'

15 = 10 x v2'

v2' = 15/10

v2' = 1.5 m/s

hence , the speed of the larger ball after the collision will 1.5m/s

Answer 2

Hi there!

Recall the conservation of momentum:

m1v1 + m2v2 = m1v1' + m2v2'

Let m1 = 5 kg ball and m2 = 10 kg ball

Since m2 is at rest, we can rewrite:

m1v1 = m1v1' + m2v2'

Plug in the given values:

5(2) = 5(-1) + 10v2'

Solve for v2':

10 = -5 + 10v2'

15 = 10v2'

15/10 = 1.5 m/s