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Patrick1998
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A bowling ball of mass 5 kg hits a wall going 7 m/s and rebounds at a speed of 2 m/s. What was the impulse applied to the bowling ball

Verified Answer :

MissKlock

The impulse applied to the bowling ball is 45 kg m/s

Explanation :

Given that,

Mass of the ball, m = 5 kg

Initial velocity of the ball, u = 7 m/s

Final speed of the ball, v = 2 m/s

We know that the impulse applied to an object is equal to the change in momentum.

\(J=\Delta p\)

\(J=p_f-p_i\)

\(J=m(v-u)\)

\(J=5\ kg(2\ m/s-7\ m/s)\)

J = -25 kgm/s

Hence, this is the required solution.

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