Question

A 15.0 kg medicine ball is thrown at a velocity of 10.0 m/s to a 60.0 kg person who is a rest on ice. The person catches the ball and immediately slides across the ice. Assume that momentum is conserved and there is no friction. Calculate the velocity of the person AND THE BALL.

Answer 1

Answer:

The velocity of ball and man after catching ball is $2m/s$

Explanation:

Since there is no friction and no external force the momentum is conserved .

Here when man catches ball then man and ball move with common velocity .

Let the final common velocity be $v_{f}$,

     Given  that  mass of the ball is  $m_{b}$ =$15kg$

    Given  that mass of man is $m=60kg$

  Initial velocity of ball = $v_{i} = 10m/s$

Now considering momentum conservation

             $p_{i} =p_{f}$

     $m_{b}v_{i} = (m_{b} +m)\times v\\v=m_{b}v_{i}/(m_{b} +m) = \frac{150}{75} =2m/s$

Answer 2

Answer:

Ball = 150kg*m/s

Explanation:

How to find momentum, p:

p = m*v    (P=MOMENTUM, M=MASS, V=VELOCITY OF BALL)

P = 15*10

P=150

For the human, momentum is always conserved, so what the momentum is before the ball is thrown must equal after the ball is thrown:

pBEFORE= pAFTER

mv = mv

(15)(10) = (60)(v)    SOLVE FOR V

150 = 60v

v = 5/3