A 50-kg soccer player jumps vertically upwards and heads the 0.45-kg ball as it is descending vertically with a speed of 30 m/s.

Question

(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?


m/s

(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.)

Answer 1

momentum is conserved, and energy in an elastic collision

up is positive

(50 * 4.0) - (.45 * 30) = (50 p) + (.45 b)
... 186.5 = 50 p + .45 b

1/2 [(50 * 4.0^2)+(.45 * 30^2)] =
... 1/2 [(50 * p^2)+(.45 * b^2)]
... 1205 = 50 p^2 + .45 b^2

solve the system of equations for b
... using substitution

(b) a = (b - -30) / .021 m/s^2

Answer 2

a. Given:
M1 = 50kg, V1 = 4 m/s.
M2 = 0.45 kg, V2 = -30 m/s.

V3 = Velocity of M1 after colliding.
V4 = Velocity of M2 after colliding.

Momentum before = Momentum after
M1*V1 + M2*V2 = M1*V3 + M2*V4.
50*4 + 0.45*(-30) = 50*V3 + 0.45*V4.
Eq1: 50V3 + 0.45V4 = 186.5.

V3 = (V1(M1-M2) + 2M2*V2)/(M1+M2).
V3 = (4(50-0.45) - 13.5)/50.45),
V3 = (198.2 - 13.5)/50.45 = 3.66 m/s.

In Eq1, replace V3 with 3.66 and solve for V4.

Answer 3

Correction:
V3 = (4(50-0.45) - 27)/(50.45),
V3 = (198.2 - 27)/50.45 = 3.39 m/s.

b. a = (V4-V2)/0.021