Question

A 69.5-kg person throws a 0.0445-kg snowball forward with a ground speed of 35.0 m/s. A second person, with a mass of 56.5 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 3.00 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. (Take the direction the snowball is thrown to be the positive direction. Indicate the direction with the sign of your answer.)

Answer 1

The velocity of the first person is 2.9795 m/s and the velocity of the second person is 0.0275 m/s

Step-by-step explanation:

First, we are going to study the interacción between the ball and the first person. In every case the linear momentum is conserved so:

`m_1v_(i1) +m_bv_(ib)=m_1v_(f1)+m_bv_(b)`

Where
`m_1` is the mass of the first person,
`m_b` is the mass of the ball,
`v_(i1)` and
`v_(f1)` are the initial and final velocities of the first person,
`v_(ib)` is the initial velocities of the ball and
`v_(b)` is the velocity of the ball once it is throws.

So, if we replace the values and solve for
`v_(f1)`, we get:

`69.5Kg(3m/s)+(0.0445Kg)(3m/s)=(69.5kg)v_(f1)+(0.0445)(35m/s)`

`208.6335=69.5v_(f1)+1.5575`

`(208.6335-1.5575)/(69.5)=v_(f1)\\2.9795=v_(f1)`

Therefore, the velocity of the first person after the snowball is exchanged is 2.9795 m/s

Now, we are going to study the interaction between the ball and the second person, we get:

`m_2v_(i2) +m_bv_(b)=m_2v_(f2)+m_bv_(fb)`

Where
`m_2` is the mass of the second person,
`v_(i2)` is the initial velocity of the second person,
`v_(f2)` is the final velocity of the second person and
`v_(fb)` is the final velocity of the ball.

`v_(f2)` is equal to
`v_(fb)`, so if we replace the values and solve for
`v_(f2)`, we get:

`(56.5kg)(0 m/s)+(0.0445)(35)=(56.5Kg)v_(f2)+(0.0445Kg)v_(f2)`

`1.5575=(56.5445)v_(f2)`

`(1.5575)/(56.5445) =v_(f2)\\0.0275=v_(f2)`

Finally, the the velocity of the second person after the snowball is exchanged is 0.0275 m/s