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A 97 kg man holding a 0.365 kg ball stands on a frozen pond next to a wall. He throws the ball at the wall with a speed of 11.3 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 collision with the wall.

Answer :

Answer:

[tex]v = 0.085 m/s[/tex]

Explanation:

First when man throws the ball then the speed of the man is given as

[tex]m_1v_1 = m_2v_2[/tex]

[tex]97 v = 0.365 \times 11.3[/tex]

[tex]v = 0.042 m/s[/tex]

now ball will rebound after collision with the wall

so speed of the ball will be same as initial speed

now again by momentum conservation we will have

[tex]m_1v_1 + m_2v_2 = (m_1 + m_2) v[/tex]

[tex]97 \times 0.042 + 0.365 \times 11.3 = (97 + 0.365) v[/tex]

[tex]v = 0.085 m/s[/tex]

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