a 75 kg man is standing at rest on ice while holding a 4kg ball. if the man throws the ball at a velocity of 3.50 m/s forward, w

Question

1 month ago

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hat will his resulting velocity be?

1 answer:

Keith_Richards [3.2K] · Answer 1 · 2026-03-22T04:50:30+03:00

Answer:

The resulting velocity for him will be 0.187 m/s in reverse direction.

Explanation:

Given:

The mass of the man is, $M=75\ kg$

The mass of the ball is, $m=4\ kg$

The initial velocity of the man is, $u_m=0\ m/s(rest)$

The initial velocity of the ball is, $u_b=0\ m/s(rest)$

The final velocity of the ball is, $v_b=3.50\ m/s$

The final velocity of the man is, $v_m=?\ m/s$

To determine this scenario, we employ the principle of momentum conservation.

This principle states that the total initial momentum equals the total final momentum.

Momentum is calculated by multiplying mass by velocity.

Initial momentum = Initial momentum of the man and the ball

Initial momentum = $Mu_m+mu_b=75\times 0+4\times 0 =0\ Nm$

Final momentum = Final momentum of the man and the ball

Final momentum = $Mv_m+mv_b=75\times v_m+4\times 3.50 =75v_m+14$

Hence, the total initial momentum equals the total final momentum

$0=75v_m+14\\\\75v_m=-14\\\\v_m=\frac{-14}{75}\\\\v_m=-0.187\ m/s$

The negative sign indicates that the man moves backward.

Thus, his final velocity ends up being 0.187 m/s backward.