Explanation:
Momentum is conserved.
a) In the first scenario, Olaf and the ball have the same final velocity.
mu = (M + m) v
(0.400 kg) (10.9 m/s) = (70.2 kg + 0.400 kg) v
v = 0.0618 m/s
b) In the second scenario, the ball has a final velocity of 8.10 m/s in the opposite direction.
mu = mv + MV
(0.400 kg) (10.9 m/s) = (0.400 kg) (-8.10 m/s) + (70.2 kg) v
v = 0.108 m/s
a) After Olaf catches the ball, Olaf and the ball will move at a speed of 0.062 m/s.
b) The speed of Olaf after the ball bounces off his chest is 0.11 m/s.
a) We can find the speed of Olaf and the ball by conservation of linear momentum.
[tex] p_{i} = p_{f} [/tex]
[tex] m_{o}v_{i_{o}} + m_{b}v_{i_{b}} = m_{o}v_{f_{o}} + m_{b}v_{f_{b}} [/tex]
Where:
[tex] m_{o}[/tex]: is the mass of Olaf = 70.2 kg
[tex] m_{b}[/tex]: is the mass of the ball = 0.400 kg
[tex] v_{i_{o}}[/tex]: is the intial speed of Olaf = 0 (he is at rest)
[tex]v_{i_{b}}[/tex]: is the initial speed of the ball = 10.9 m/s
[tex] v_{f_{o}}[/tex] and [tex]v_{f_{b}} [/tex]: are the final speed of Olaf and the final speed of the ball, respetively.
Since Olaf catches the ball, we have that [tex] v_{f_{o}} = v_{f_{b}} = v [/tex], so:
[tex] m_{o}v_{i_{o}} + m_{b}v_{i_{b}} = v(m_{o} + m_{b}) [/tex]
We will take the direction of motion of the ball to the right side, and this will be the positive x-direction.
By solving for "v" we have:
[tex] v = \frac{m_{o}v_{i_{o}} + m_{b}v_{i_{b}}}{m_{o} + m_{b}} = \frac{70.2 kg*0 + 0.400 kg*10.9 m/s}{70.2 kg + 0.400 kg} = 0.062 m/s [/tex]
Hence, Olaf and the ball will move at a speed of 0.062 m/s.
b) The final speed of Olaf after the collision can be calculated, again with conservation of linear momentum.
[tex] m_{o}v_{i_{o}} + m_{b}v_{i_{b}} = m_{o}v_{f_{o}} + m_{b}v_{f_{b}} [/tex]
In this case, since the ball hits Olaf and bounces off his chest, we have that [tex] v_{f_{o}} \neq v_{f_{b}}[/tex]
[tex] 0.400 kg*10.9 m/s = 70.2 kg*v_{f_{o}} + 0.400 kg*(-8.10 m/s) [/tex]
The minus sign of the speed of the ball is because it moves to the negative direction of motion after the collision.
[tex] v_{f_{o}} = \frac{0.400 kg*10.9 m/s + 0.400 kg*8.10 m/s}{70.2 kg} = 0.11 m/s [/tex]
Therefore, the speed of Olaf after the collision will be 0.11 m/s in the positive x-direction.
You can see another example of conservation of linear momentum here: https://brainly.com/question/22257322?referrer=searchResults
I hope it helps you!
