A 35-kg girl is standing near and to the left of a 43-kg boy on the frictionless surface of a frozen pond. The boy throws a 0.75-kg ice ball to the girl with a horizontal speed of 6.2 m/s. What are the velocities of the boy and the girl immediately after the girl catches the ice ball?

Question

Corbin Reeves

Physics

25 June, 21:11

A 35-kg girl is standing near and to the left of a 43-kg boy on the frictionless surface of a frozen pond. The boy throws a 0.75-kg ice ball to the girl with a horizontal speed of 6.2 m/s. What are the velocities of the boy and the girl immediately after the girl catches the ice ball?

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Chace · Answer 1

The velocity of the boy is 0.11 m/s to the right.

The velocity of the girl is 0.13 m/s to the left.

Explanation:

Hi there!

The momentum of the system is conserved. That means that the initial momentum is equal to the final momentum (after throwing the ice ball for the boy and after catching the ball for the girl).

Let's calculate the momentum of the system boy-ball:

The equation of the momentum is the following:

p = m · v

Where:

p = momentum.

m = mass of the object.

v = velocity.

Initially, the ball and the boy are at rest, so that the momentum of the system boy-ball is zero:

initial momentum = mib · vib + mb · vb

Where:

mib = mass of the ice ball.

vib = velocity of the ice ball.

mb = mass of the boy.

vb = velocity of the boy.

The initial momentum is zero because both velocities are zero.

After throwing the ball, the momentum of the system will be:

final momentum = mib · vib + mb · vb

final momentum = 0.75 kg · ( - 6.2 m/s) + 43 kg · vb (Let's consider the left as negative).

Since final momentum = initial momentum

0 kg · m/s = - 4.65 kg · m/s + 43 kg · vb

Solving for vb:

4.65 kg · m/s/43 kg = vb

vb = 0.11 m/s

The velocity of the boy after releasing the ball is 0.11 m/s. Since there is no friction, this velocity is constant. When the girl catches the ball, the velocity of the boy will be 0.11 m/s (to the right).

Now let's do the same for the system girl-ball.

initial momentum = mg · vg + mib · vib

initial momentum = 35 kg · 0 m/s + 0.75 kg · (-6.2 m/s)

initial momentum = - 4.65 kg · m/s

The final momentum is the momentum of the girl plus the ball:

final momentum = (mg + mb) · vgb

final momentum = (35 kg + 0.75 kg) · vgb

Since initial momentum = final momentum

-4.65 kg · m/s = (35 kg + 0.75 kg) · vgb

-4.65 kg · m/s / (35 kg + 0.75 kg) = vgb

vgb = - 0.13 m/s

The velocity of the girl after catching the ball is - 0.13 m/s (0.13 m/s to the left).