Question

In an inelastic collision a 2.5 kg ball moving at 7.5 m/s is caught by a 70kg man while the man is standing on ice. What is the velocity of the ball and the man?

Answer 1

The velocity of the ball and the man is 0.259 m/s

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, in an isolated system, the total momentum before and after the collision must be conserved. Therefore, for the ball-man system, we can write:

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

where:

$m_1 = 2.5 kg$ is the mass of the ball

$u_1 = 7.5 m/s$ is the initial velocity of the ball

$m_2 = 70 kg$ is the mass of the man

$u_2 = 0$ is the initial velocity of the man

$v$ is the final velocity of the man and the ball after the collision

Re-arranging the equation and substituting the values, we find the final velocity:

$v=\frac{m_1 u_1}{m_1+m_2}=\frac{(2.5)(7.5)}{2.5+70}=0.259 m/s$

So, the man and the ball slides on the ice at 0.259 m/s.

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