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A 0.5-kg ball moving at 5 m/s strikes a wall and rebounds in the opposite direction with a speed of 2 m/s. If the impulse occurs for a time duration of 0.01 s, then the average force (magnitude only) acting upon the ball is ____ Newtons.

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Given that:

mass of the ball (m) = 0.5 Kg ,

ball strikes the wall (v₁) = 5 m/s ,

rebounds in opposite direction (v₂) = 2 m/s,

time duration (t) = 0.01 s,

Determine the force (F) = ?

We know that from Newton's II law,

F = m. a Newtons

(velocity acting in opposite direction, so a = ( (v₁ + v₂)/t

= m × (v₁ + v₂)/t

= 0.5 × (5 + 2)/0.01

= 350 N

The force acting up on the ball is 350 N

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