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sukhopar [10]
3 years ago
12

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.
Physics
1 answer:
Elanso [62]3 years ago
3 0

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.

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