Question

A 62.4 kg ice skater moving to the right with a velocity of 2.67 m/s throws a 0.198 kg snow ball to the right with a velocity of 39.7 m/s relative to the ground. t is the velocity of the ice skater after throwing the snowball? Disregard the friction between the skates and the ice. Answer in units of m/s

A second skater initially at rest with a mass of 58.4 kg catches the snowball. What is the velocity of the second skater after catching the snowball in a perfectly inelastic collision? Answer in units of m/s

Answer 1

1) Final velocity of the ice skater is 2.55 m/s

2) Final velocity of the second skater is 0.134 m/s

Explanation:

1)

We can solve this problem by using the principle of conservation of momentum. In fact, the total momentum of the ice skater-snowball system must be conserved before and after the shot, so we can write:

$p_i = p_f\\(m_1 +m_2)u = m_1 v_1 + m_2 v_2$

where:

$m_1 = 62.4 kg$ is the mass of the ice skater

$m_2 = 0.198 kg$ is the mass of the ball

$u = 2.67 m/s$ is the initial velocity of the skater and the ball

$v_1$ is the final velocity of the skater

$v_2 = 39.7 m/s$ is the final velocity of the ball

Re-arranging the equation and solving for v1, we find the final velocity of the ice skater:

$v_2 = \frac{(m_1+m_2)u-m_2 v_2}{m_1}=\frac{(62.4+0.198)(2.67)-(0.198)(39.7)}{62.4}=2.55 m/s$

And since the sign is positive, the direction is the same as the initial direction, therefore to the right.

2)

Again, we can apply the law of conservation of momentum. This time we have

$p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v$  

where:  

$m_1 = 58.4 kg$ is the mass of the second ice skater

$u_1 = 0$ is his initial velocity

$m_2 = 0.198 kg$ is the mass of the ball

$u_2 = 39.7 m/s$ is the initial velocity of the ball

$v$ is the final combined velocity of the  skater + the ball

Re-arranging the equation and solving for v, we find:  

$v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{0+(0.198)(39.7)}{58.4+0.198}=0.134 m/s$

And the positive sign indicates their final direction is right again.

Learn more about momentum here:

brainly.com/question/7973509  

brainly.com/question/6573742  

brainly.com/question/2370982  

brainly.com/question/9484203  

#LearnwithBrainly

Answer 2

The final velocity of the first skater is 2.544 m/s and the velocity of the second skater is 0.134 m/s.

In this question, the concepts of conservation of momentum and perfectly inelastic collision are to be used.

let, m(1) = 62.4 kg , m(b) = 0.198 kg and m(2) = 58.4 kg be the masses of the first skater, the ball and the second skater respectively.

(i) the first skater throws the ball

Their initial velocities are u(1) = 2.67 m/s , u(b) =0

and the final velocities are v(1) , v(b) = 39.7 m/s

Applying conservation off momentum:

   m(1)u(1) + m(b)u(b) = m(1)v(1) + m(b)v(b)

62.4 × 2.67 + 0.198 × 0 = 62.4 × v(1) + 0.198 × 39.7

v(1) = $\frac{62.4*2.67 - 0.198*39.7}{62.4}$

v(1) = 2.544 m/s is the final speed of the first skater after throwing th ball.

(ii) The second skater catches the ball

Here the collision is perfectly inelastic, both the ball and skater move with the same velocity (v) after catching the ball.

Applying conservation of momentum

m(b)v(b) = [m(2) + m(b)] v

0.198 × 39.7 = [58.4 + 0.198] v

v = 0.134 m/s is the velocity of the second skater after catching the ball.

Learn more about elastic and inelastic collision:

https://brainly.com/question/24170708?referrer=searchResults