Cart Collision Homework: Calculate Distance Traveled After Bounce

  • Thread starter Thread starter MightyMan11
  • Start date Start date
  • Tags Tags
    Cart Momentum
AI Thread Summary
A 478.0 g cart is released from a 1.16 m high frictionless ramp at a 31.0° angle and collides with a rubber block, prompting a calculation of how far it rolls back up after bouncing. Initial calculations for the cart's velocity and momentum were performed, but the final distance traveled up the ramp was incorrectly derived due to using cosine instead of sine for the angle in the acceleration calculation. The correct approach involves using the sine function to accurately determine the distance. The participant acknowledges the error and appreciates the assistance received. The discussion emphasizes the importance of using the correct trigonometric functions in physics problems.
MightyMan11
Messages
13
Reaction score
0

Homework Statement


A 478.0 g cart is released from rest 1.16 m from the bottom of a frictionless, 31.0° ramp. The cart rolls down the ramp and undergoes a collision with a rubber block at the bottom. After the cart bounces, how far does it roll back up the ramp if the maximum force applied is 302.0 N for 22.9 ms?

Homework Equations



pf=pi+impulse
Vf2=Vi2+2ad
p=m*v

The Attempt at a Solution


First I found my initial impulse:
Vf2=Vi2+2ad
Vf2=0+2(9.8cos31)(1.16)
Vf=4.41m/s
but since its moving in the negative direction, it is -4.41m/s

p = m*v
p = (0.478)(-4.41)
=-2.11

The I calculate my impulse (the area under the graph):
0.5(302)(0.0229)
=3.46 kg m/s

Then I calculated my final impulse:
pf=pi+impulse
pf=-2.11+3.46
=1.352

I then found the distance it travels up after it gains that momentum from hitting the rubber block:
p=m*Vi
1.352=0.478*Vi
Vi=2.83

Vf2=Vi2+2ad
0=2.832+2(-9.81cos31)d
d=0.477m

However, this answer does not match my answer key...

Thanks in advanced
 
Last edited:
Physics news on Phys.org
Welcome to PF!

MightyMan11 said:
A 478.0 g cart is released from rest 1.16 m from the bottom of a frictionless, 31.0° ramp. The cart rolls down the ramp and undergoes a collision with a rubber block at the bottom. After the cart bounces, how far does it roll back up the ramp if the maximum force applied is 302.0 N for 22.9 ms?
…
First I found my initial impulse:
Vf2=Vi2+2ad
Vf2=0+2(9.8cos31)(1.16)

Hi MightyMan11! Welcome to PF! :smile:

Isn't it sin31º?
 


tiny-tim said:
Hi MightyMan11! Welcome to PF! :smile:

Isn't it sin31º?

AHHH...
Looks like it.

Thanks for the help and the welcome :smile:
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
View full post »

Similar threads

Calculating Velocity After Impulse | Momentum and Impulse Homework Help
Replies
2
Views
1K
Impulse and momentum of a cart down a ramp
Replies
1
Views
6K
Cart on an inclined ramp, with rebound
Replies
4
Views
17K
Momentum and Impulse during a collision
Replies
13
Views
5K

Hot Threads

Recent Insights

Back
Top