A 50-kg soccer player jumps vertically upwards and heads the 0.45-kg ball as it

Question

A 50-kg soccer player jumps vertically upwards and heads the 0.45-kg ball as it is descending vertically with a speed of 20 m/s. (a) If the player was moving upward with a speed of 4.0 m/s just before impact, what will be the speed of the ball immediately after the collision if the ball rebounds vertically upwards and the collision is elastic?

(b) If the ball is in contact with the player's head for 21 ms, what is the average acceleration of the ball? (Note that the force of gravity may be ignored during the brief collision time.)

Explanation / Answer

m1 = 50 kg , u1 = 4m/s , m2 =0.45 kg ,u2 = -20 m/s

From conservation of linera momentum along vertical direction

m1u1+m2u2 =m1v1+m2v2

(50*4) + (0.45*-20) = 50v1+0.45v2

191 = 50v1+0.45 v2 ... (1)

In elastic collision , e =1

v2-v1 = u1 -u2 =4+20

v2 -v1 = 24 .... (2)

By solving (1) and (2), we get

v1 = 3.57 m/s

v2 = 27.57 m/s

(b) t =21 ms

impulse =Force*time = change in momentum

F*t = m (v2-u2)

F*0.021 = 0.45(27.57+20)

F = 1019.36 N

F =ma

a = F/m = 1019.36/0.45

a = 2265.24 m/s^2


Force = mass*acceleration

acceleration = 1450 m/s^2

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