Question

A 70.0-kg person throws a 0.0430-kg snowball forward with a ground speed of 32.0 m/s. A second person, with a mass of 58.5 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 3.30 m/s, and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged

Answer 1

The velocities of the skaters are
`v_(1) = 3.280\,(m)/(s)` and
`v_(2) = 0.024\,(m)/(s)`, respectively.

Step-by-step explanation:

Each skater is not under the influence of external forces during process, so that Principle of Momentum Conservation can be used on each skater:

First skater

`m_(1) \cdot v_(1, o) = m_(1) \cdot v_(1) + m_(b)\cdot v_(b)` (1)

Second skater

`m_(b)\cdot v_(b) = (m_(2)+m_(b))\cdot v_(2)` (2)

Where:

`m_(1)` - Mass of the first skater, in kilograms.

`m_(2)` - Mass of the second skater, in kilograms.

`v_(1,o)` - Initial velocity of the first skater, in meters per second.

`v_(1)` - Final velocity of the first skater, in meters per second.

`v_(b)` - Launch velocity of the meter, in meters per second.

`v_(2)` - Final velocity of the second skater, in meters per second.

If we know that
`m_(1) = 70\,kg`,
`m_(b) = 0.043\,kg`,
`v_(b) = 32\,(m)/(s)`,
`m_(2) = 58.5\,kg` and
`v_(1,o) = 3.30\,(m)/(s)`, then the velocities of the two people after the snowball is exchanged is:

By (1):

`m_(1) \cdot v_(1, o) = m_(1) \cdot v_(1) + m_(b)\cdot v_(b)`

`m_(1)\cdot v_(1,o) - m_(b)\cdot v_(b) = m_(1)\cdot v_(1)`

`v_(1) = v_(1,o) - \left((m_(b))/(m_(1)) \right)\cdot v_(b)`

`v_(1) = 3.30\,(m)/(s) - \left((0.043\,kg)/(70\,kg)\right)\cdot \left(32\,(m)/(s) \right)`

`v_(1) = 3.280\,(m)/(s)`

By (2):

`m_(b)\cdot v_(b) = (m_(2)+m_(b))\cdot v_(2)`

`v_(2) = (m_(b)\cdot v_(b))/(m_(2)+m_(b))`

`v_(2) = ((0.043\,kg)\cdot \left(32\,(m)/(s) \right))/(58.5\,kg + 0.043\,kg)`

`v_(2) = 0.024\,(m)/(s)`