A 55-kg soccer player jumps vertically upwards and heads the 0.45-kg ball as it

Question

A 55-kg soccer player jumps vertically upwards and heads the 0.45-kg ball as it is descending vertically with a speed of 27 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 24 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

here,

mass of player , m1 = 55 kg

mass of ball , m2 = 0.45 kg

initial speed of ball l, u1 = - 27 m/s

initial speed of player , u2 = 4 m/s

let the final speeds be v1 and v2

using conservation of momentum

m1 * u1 + m2 * u2 = m1 * v1 + m2 * v2

55 * 4 - 0.45 * 27 = 55 * v1 + 0.45 * v2 ....(1)

and

using conservation of kinetic energy

0.5 * m1 * u1^2 + 0.5 * m2 * u2^2 = 0.5 * m1 * v1^2 + 0.5 * m2 * v2^2

55 * 4^2 + 0.45 * 27^2 = 55 * v1^2 + 0.45 * v2^2 ....(2)

from (1) and (2)

v1 = 3.5 m/s

v2 = 34.5 m/s

a)

the final speed of ball is 34.5 m/s upwards

b)

time taken , t = 24 ms = 0.024 s

the average acceleration of the ball , a = ( v2 - u2)/t

a = ( 34.5 + 27)/0.024 m/s^2

a = 2562.5 m/s^2

the average accelration of the ball is 2562.5 m/s^2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.