Calculation of Velocity in reference to circular motion

[Kdoll1]

Homework Statement

1. A boy with a mass of 50kg jumps off a 30kg boat, causing the boat to move to the right at 2 m/sec. What did the boy jump with a velocity of?
a.-1.2 m/sec
b.3.3 m/sec
c.1.1 m/sec
d.0.2 m/sec
e.-3.3 m/sec

2. A 2kg cart moving in a straight line hits a piece of modeling clay that has a mass of 1kg and is at rest. After the collision, the cart and clay stick together and move at 1m/sec to the left. The initial velocity of the cart is ___m/sec to the left.
a.1
b.1.5
c.2
d.2.5
e.-0.33

Homework Equations

m1*v2/r=g*m1m2/r2...all of those numbers are subscripts

The Attempt at a Solution

1. v=sq. root of Gm/r
v=(6.67*10^-11)20/2=
1.334*10 to the second

2. 1/3


I am beyond confused...physics does not click in my head at all.

 

[alphysicist]

Hi Kdoll1,

Homework Statement

1. A boy with a mass of 50kg jumps off a 30kg boat, causing the boat to move to the right at 2 m/sec. What did the boy jump with a velocity of?
a.-1.2 m/sec
b.3.3 m/sec
c.1.1 m/sec
d.0.2 m/sec
e.-3.3 m/sec

2. A 2kg cart moving in a straight line hits a piece of modeling clay that has a mass of 1kg and is at rest. After the collision, the cart and clay stick together and move at 1m/sec to the left. The initial velocity of the cart is ___m/sec to the left.
a.1
b.1.5
c.2
d.2.5
e.-0.33

Homework Equations

m1*v2/r=g*m1m2/r2...all of those numbers are subscripts

This equation describes the motion of satellites, which would not apply here, and the problems you posted do not deal with circular motion. Instead, the problems deal with conservation of momentum, so what is going wrong is you are using the wrong approach. So read the conservation of momentum section and try them again, and see if you can get the answer.

 

[baba89]

Can someone please explain this to me?

Sure, let's break it down step by step.

1. First, let's define some variables:
- m1 = mass of the boy (50kg)
- m2 = mass of the boat (30kg)
- v1 = velocity of the boy before jumping (unknown)
- v2 = velocity of the boy after jumping (unknown)
- v3 = velocity of the boat after the boy jumps (2 m/sec)

2. Now, let's use the conservation of momentum principle to solve for v2:
- Before the jump, the total momentum of the system (boy + boat) is 0, since they are at rest.
- After the jump, the total momentum of the system is still 0, since there are no external forces acting on it.
- Therefore, we can use the equation: m1v1 + m2v2 = m1v2 + m2v3
- Plugging in the values we know, we get: (50kg)(v1) + (30kg)(0) = (50kg)(v2) + (30kg)(2 m/sec)
- Simplifying, we get: 50v1 = 50v2 + 60
- Subtracting 50v2 from both sides, we get: 50v1 - 50v2 = 60
- Factoring out 50, we get: 50(v1 - v2) = 60
- Dividing both sides by 50, we get: v1 - v2 = 1.2
- Adding v2 to both sides, we get: v1 = 1.2 + v2
- So, the boy jumped with a velocity of 1.2 + v2 m/sec, where v2 is the velocity of the boy after the jump.

3. We know that the boy jumps off the boat, so his velocity is in the opposite direction of the boat's velocity (to the left). Therefore, v2 must be negative. Now, let's look at the answer choices:
a. -1.2 m/sec
b. 3.3 m/sec
c. 1.1 m/sec
d. 0.2 m/sec
e. -3.3 m/sec
- We can eliminate options b, c, and d, since they are all positive and we know v