Two ice skaters want to prove conservation of momentum. Skater A has a mass of 72 KG, and skater B has a mass of 48 KG. they are originally at rest. they hold hands and face each other. they then push off each other and move in opposite directions. skater b moves with a velocity of 5MS. what must the velocity of skater A be in the opposite direction after the push in order to prove conservation of momentum?

Answer 1

To prove conservation of momentum, we need to ensure that the total momentum before the push is equal to the total momentum after the push.

Before the push, both skaters are at rest, so the total initial momentum is:

\[ \text{Total initial momentum} = 0 \]

After the push, skater B (mass = 48 kg) moves with a velocity of 5 m/s in one direction. The momentum of skater B after the push is:

\[ \text{Momentum of skater B} = \text{mass} \times \text{velocity} = 48 , \text{kg} \times 5 , \text{m/s} = 240 , \text{kg m/s} \]

Let \( v_A \) be the velocity of skater A (mass = 72 kg) in the opposite direction. The momentum of skater A after the push is:

\[ \text{Momentum of skater A} = 72 , \text{kg} \times v_A \]

Now, we apply the conservation of momentum, setting the total initial momentum equal to the total final momentum:

\[ 0 = (72 , \text{kg} \times v_A) + (48 , \text{kg} \times 5 , \text{m/s}) \]

Rearranging the equation yields:

\[ 72 , v_A + 240 = 0 \]

Solving for \( v_A \):

\[ 72 , v_A = -240 \] \[ v_A = -\frac{240}{72} \] \[ v_A = -3.33 , \text{m/s} \]

The negative sign indicates that skater A moves in the opposite direction to skater B. Therefore, skater A must move with a velocity of approximately \( 3.33 , \text{m/s} \) in the opposite direction after the push in order to prove conservation of momentum.