A ball of mass 0.220 kg that is moving with a speed of 7.5 m/s collides head-on and elastically with another ball initially at rest. Immediately after the collision, the incoming ball bounces backward with a speed of 3.8 m/s. (a) Calculate the velocity of the target ball after the collision, and (b) the mass of the target ball.

The velocity of the target ball after the collision is 3.7m/s

The mass of the target ball is 0.22kg

To calculate the velocity of the target ball after the collision, we will use the law of conservation of momentum expressed according to the formula

are the initial velocities of the object

are the final velocities

a) Given the following parameters

Substitute the given parameters into the formula to have:

Hence the velocity of the target ball after the collision is 3.7m/s

b) To get the mass of the target ball, we will use the formula;

Substitute the masses and velocities

Hence the mass of the target ball is 0.22kg

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Explanation:

Given that,

Mass of ball 1

M1= 0.22kg

Initial Speed of ball 1

U1 = 7.5m/s

Initially ball 2 is at rest

U2 = 0m/s

Let mass of ball 2 be M2

Velocity of ball 1 after collision

V1 = 3.8m/s

A. Velocity of ball 2 after collision

Let the velocity of ball two after collision be V2

Using conservation of energy of momentum, then the formula reduces to

U1—U2 = —(V1 —V2)

Note that V1 is moving in the negative direction.

Check attachment

7.5—0 = —(—3.8 —V2)

7.5 = 3.8+V2

V2= 7.5—3.8

V2 = 3.7 m/s

B. Mass of ball 2= M2

Using conservation of momentum

P(before) = P(after)

M1•U1 + M2•U2 = —M1•V1 + M2•V2

0.22×7.5 + 0= —0.22 × 3.8 + 3.7•M2

1.65 = —0.836 + 3.7•M2

3.7•M2 = 1.65+0.836

M2 = 2.486/3.7

M2 = 0.67 kg

answer

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