A 70 kg man holding a 0

.
857 kg ball stands
on a frozen pond next to a wall. He throws
the ball at the wall with a speed of 11
.
9 m
/
s
(relative to the ground) and then catches the
ball after it rebounds from the wall.
How fast is he moving after he catches the
ball? Ignore the projectile motion of the ball,
and assume that it loses no energy in its colli-
sion with the wall.
Answer in units of m
/
s

To find the man's speed after catching the ball, we need to apply the principle of the conservation of momentum. In this case, momentum is defined as the mass of an object multiplied by its velocity.

The initial momentum of the man and the ball before throwing the ball can be calculated by summing up the individual momenta:

Initial momentum = (mass of the man × velocity of the man) + (mass of the ball × velocity of the ball)

Given:
- Mass of the man (m1) = 70 kg
- Mass of the ball (m2) = 0.857 kg
- Velocity of the man (v1) = 0 m/s (since he is initially at rest)
- Velocity of the ball (v2) = 11.9 m/s (relative to the ground)

Initial momentum = (70 kg × 0 m/s) + (0.857 kg × 11.9 m/s)
Initial momentum = 0 + 10.1683 kg·m/s = 10.1683 kg·m/s

According to the conservation of momentum, the total momentum before throwing the ball should be equal to the total momentum after catching the ball. Therefore, the final momentum (momentum after catching the ball) should also be 10.1683 kg·m/s.

Since the man catches the ball, the velocity of the ball after rebounding from the wall will be opposite in direction to its initial velocity. So the velocity of the ball after rebounding will be -11.9 m/s.

Final momentum = (mass of the man × final velocity of the man) + (mass of the ball × final velocity of the ball)

To find the final velocity of the man after catching the ball (v3), we can rearrange the equation:

(mass of the man × final velocity of the man) = Final momentum - (mass of the ball × final velocity of the ball)

v3 = (Final momentum - (mass of the ball × final velocity of the ball)) ÷ mass of the man

Substituting the given values:
v3 = (10.1683 kg·m/s - (0.857 kg × -11.9 m/s)) ÷ 70 kg

v3 = (10.1683 kg·m/s + 10.2083 kg·m/s) ÷ 70 kg

v3 = 20.3766 kg·m/s ÷ 70 kg

v3 ≈ 0.291 kg·m/s (rounded to three decimal places)

Therefore, the man's speed after catching the ball is approximately 0.291 m/s.