Need help with this physics question!

  • Thread starter bobo2210
  • Start date
  • #1
7
0
If a ball of mass of 0.1kg is placed at the end of the compressed spring
(k=4000N/m). If the spring is compressed 1cm...
a. What will the kinetic energy of the ball be?
b. How fast is it moving?
c. What is its momentum?
d. What was the force on the ball from the spring?
e. If it took 0.01s to go from compression to ‘launch’, what is the impulse?
f. What was the power?
g. If this ball was on the bottom of a frictionless roller coaster track, what is
the maximum height that the loop could be?
h. What is the speed of the ball at the bottom of the loop?
i. If this 0.1kg mass collides with a 1.0kg mass that was at rest, what is the
momentum of the system after the collision? (this is an elastic collision)
j. What are the velocities (magnitude and direction) of 0.1kg and 1.0kg
masses

What I got for (a)
is KE=1/2*mv^2
KE= 1/2*(0.1kg)*(0.01m/s)= 0.0005J
b) the ball is moving slow
then I didn't know how to find the momentum
 

Answers and Replies

  • #2
CWatters
Science Advisor
Homework Helper
Gold Member
10,541
2,309
Where did you get v2=0.01m/s from ?

If a ball of mass of 0.1kg is placed at the end of the compressed spring
(k=4000N/m). If the spring is compressed 1cm...
a. What will the kinetic energy of the ball be?

I assume the full question says

"If the spring is compressed 1cm and then released..
a. What will the kinetic energy of the ball be?"

Are you familiar with the equation for the amount of energy stored in a compressed spring..

PEs = 0.5kx2
 
  • #3
7
0
Yes this is the equation for the change in potential energy. I solved this and got 0.2 for the change in potential energy. Do I have to plug it into the equation ΔPE+ΔKE= K ?
 
  • #4
CWatters
Science Advisor
Homework Helper
Gold Member
10,541
2,309
Yes. You can assume all the PE in the spring ends up in the ball as KE. So the answer to a) is just 0.2 Joules.
 
  • #5
7
0
Okay so to find the velocity for B I solved for v form the equation KE=1/2MV2
and got 4m/s2. Knowing the velocity, I plugged it into the momentum equation to solve for momentum and got 0.4kg8m/s. Is this the correct way to solve for b and c?
 
  • #6
CWatters
Science Advisor
Homework Helper
Gold Member
10,541
2,309
Yes, right approach but double check your answer.
 

Related Threads on Need help with this physics question!

  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
17
Views
1K
  • Last Post
Replies
4
Views
901
Replies
4
Views
1K
Top