A 67 kg soccer player jumps vertically upwards and heads the 0.45 kg ball as it

Question

A 67 kg soccer player jumps vertically upwards and heads the 0.45 kg ball as it is descending vertically with a speed of 29 m/s. 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?
___ m/s

If the ball is in contact with the player's head for 20 ms, what is the average acceleration of the ball? (Note that the force of gravity may be ignored during the brief collision time.)
___ m/s2

Explanation / Answer

as there is no external force b/w ball and head so the momentum will be conserve

we can write

m1xu1 + m2xu2 = m1xv1+m2xv2

m1*(u1-v1) = m2*(v2-u2)........(1)

by energy conservation

1/2 x m1x u12 + 1/2x m2x u22 = 1/2 xm1xv12 + 1/2xm2xv22

m1x(u12 - v12) = m2x(v22-u22).....(2)

from 1 &2

u1 + v1 = u2+v2

v2 = u1 - u2 + v1........(3)

put 3 in 1

m1x(u1-v1) = m2x(u1 - u2 + v1 - u2)

v1 = u1x(m1-m2)/(m1+m2) + 2xm2xu2/(m1+m2)

and

v2 = u2*(m2-m1)/(m1+m2) + 2*m1*u1/(m1+m2)

given

m1 = 67 kg ; m2 = 0.45 kg; u1 = 4 m/s ; u2 = -29 m/s  

so v2 = (-29) x (-66.55)/67.45 + 2 x 67 x 4 /(67.45)

v2 = 36.56 m/s

part b) <a>avg = (v2-u2)/t = 36.56 - (-29) / (20x 10-3 )

= 3278 m/s2

answer

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