A 67.0-kg person throws a 0.0410-kg snowball forward with a ground speed of 29.0

Question

A 67.0-kg person throws a 0.0410-kg snowball forward with a ground speed of 29.0 m/s. A second person, with a mass of 60.0 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 2.20 m/s, and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged? Disregard the friction between the skates and the ice.

thrower    
Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. m/s catcher     A 67.0-kg person throws a 0.0410-kg snowball forward with a ground speed of 29.0 m/s. A second person, with a mass of 60.0 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 2.20 m/s, and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged? Disregard the friction between the skates and the ice.

Explanation / Answer

By the law of momentum conservation:-
For 1st person:-
(m1+m2)u = m1v1+m2v2
=(67+0.041) x 2.2

= 67 x v1 + 0.041 x 29
=67v1 = 147.4902 - 1.189
=v1 = 146.3012/67 = 2.1836 m/s

For 2nd person:-
m1u1+m2u2 = (m1+m2) x v
=>0+0.041 x 29 = (60+0.041) x v
=>v = 1.189/60.041 = 0.0198m/s

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