Question

A softball of mass 0.220 kg that is moving with a speed of 5.5 m/s (in the positive direction) collides head-on and elastically with another ball initially at rest. Afterward the incoming softball bounces backward with a speed of 3.9 m/s. (a) Calculate the velocity of the target ball after the collision. (b) Calculate the mass of the target ball.

Answer 1

Answer:

The velocity and mass of the target ball are 1.6 m/s and 1.29 kg.

Explanation:

Given that,

Mass of softball = 0.220 kg

Speed = 5.5 m/s

(a). We need to calculate the velocity of the target ball

Using conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

$0.220\times5.5+m_{2}\times0=0.220\times(-3.9)+m_{2}v_{2}$

$1.21=-0.858+m_{2}v_{2}$

$m_{2}v_{2}=2.068$....(I)

The velocity approach is equal to the separation of velocity

$u_{1}-u_{2}=v_{2}-v_{1}$

$5.5-0=v_{2}-(-3.9)$

$v_{2}=1.6\ m/s$

(b). We need to calculate the mass of the target ball

Now, Put the value of v₂ in equation (I)

$m_{2}\times1.6=2.068$

$m_{2}=\dfrac{2.068}{1.6}$

$m_{2}=1.29\ kg$

Hence, The velocity and mass of the target ball are 1.6 m/s and 1.29 kg.