Question

A(n) 75.9 kg person throws a(n) 0.059 kg
snowball forward with a ground speed of
31.5 m/s. A second person, of mass 65 kg,
catches the snowball. Both people are on
skates. The first person is initially moving
forward with a speed of 2.36 m/s, and the
second person is initially at rest.
What is the velocity of the first person im-
mediately after the snowball is thrown? Dis-
regard friction between the skates and the ice.
Answer in units of m/s

Answer 1

Task:

A 75.9 kg person throws a 0.059 kg snowball forward with a ground speed of 31.5 m/s. A second person, of mass 65 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 2.36 m/s, and the second person is initially at rest. What is the velocity of the first person immediately after the snowball is thrown? Disregard friction between the skates and the ice. Answer in units of m/s

Solution:

$$M _ {1} = 75.9 \text{ kg},$$ $$V _ {1} = 2.36 \frac {\text{m}}{\text{s}}$$ $$m _ {1} = 0.059 \text{ kg},$$ $$v _ {1} = 31.5 \frac {\text{m}}{\text{s}},$$ $$M _ {2} = 65 \text{ kg},$$ $$V _ {2} = 0 \frac {\text{m}}{\text{s}}$$ $$(M _ {1} + m _ {1}) V _ {1} = M _ {1} V _ {1} ^ {\prime} + m _ {1} v _ {1}$$ $$V _ {1} ^ {\prime} = \frac {(M _ {1} + m _ {1}) V _ {1} - m _ {1} v _ {1}}{M _ {1}} = \frac {(75.9 \text{ kg} + 0.059 \text{ kg}) \cdot 2.36 \frac {\text{m}}{\text{s}} - 0.059 \text{ kg} \cdot 31.5 \frac {\text{m}}{\text{s}}}{75.9 \text{ kg}} = 2.337 \frac {\text{m}}{\text{s}}$$

Answer:

$$V _ {1} ^ {\prime} = 2.337 \frac {\text{m}}{\text{s}}$$