Question

Sally finds herself stranded on a frozen pond so slippery that she can't stand up or walk on it. To save herself, she throws one of her heavy boots horizontally, directly away from the closest shore. Sally's mass is 60 kg, the boot's mass is 5 kg, and Sally throws the boot with speed equal to 30 m/s. a) What is Sally's speed immediately after throwing the boot? b) Where is the center of mass of the Sally-boot system, relative to where she threw the boot, after 10 s? c) How long does it take Sally to reach the shore, a distance of 30 m away from where she threw the boot? For all parts, assume the ice is frictionless.

Answer 1

Answer:

-2.5 m/s

25 m

12 seconds

Explanation:

Here, as Sally's height and shore distance are not given we are assuming that she slid her boot.

$m_1$ = Sally's mass

$m_2$ = Boot mass

$v_1$ = Sally's velocity

$v_2$ = Boot's velocity

$m_1v_1+m_2v_2=0\\\Rightarrow v_1=-\frac{m_2v_2}{m_1}\\\Rightarrow v_1=-\frac{5\times 30}{60}\\\Rightarrow v_1=-2.5\ m/s$

Velocity of Sally is -2.5 m/s

Distance = Speed × Time

$Distance=2.5\times 10=25\ m$

Sally traveled 25 m in 10 seconds

Time = Distance / Speed

$Time=\frac{30}{2.5}=12\ s$

If sally was 30 m away from the shore then it would take her 12 seconds to get to the shore at 2.5 m/s.

Answer 2

The answers to the following questions are as follows:

1. -2.5 m/s
2. 25 m
3. 12 seconds

How to calculate speed?

According to this question, the following details are given about Sally:

• m1 = Sally's mass = 60kg
• m2 = Boot mass = 5kg
• v1 = Sally's velocity
• v2 = Boot's velocity = 30m/s

M1V1 + M2V2 = 0

V1 = M2V2/M1

V1 = 5 × 30/60

V1 = 2.5m/s

Sally's velocity = 2.5m/s.

Distance = Speed × Time

Distance = 2.5m/s × 10s

Distance = 25m

Time = Distance / Speed

Time = 30/2.5 = 12s

Therefore, if sally was 30 m away from the shore then it would take her 12 seconds to get to the shore at 2.5 m/s.

Learn more about distance at: https://brainly.com/question/26711747