Collision with projectile and block attached to spring

AI Thread Summary
The discussion revolves around a physics problem involving a projectile colliding with a block attached to a spring. The impulse delivered to the block is calculated as 0.4 N⋅s, with an average force of 100 N during the 4 ms collision. The collision is determined to be inelastic, as there is a significant loss of kinetic energy, approximately 60%. Participants emphasize the importance of vector direction in momentum calculations and suggest using diagrams for clarity. The conversation highlights the need for precise definitions and careful handling of vector quantities in solving such problems.
x24759
Messages
6
Reaction score
0

Homework Statement


A projectile of mass m = 50g traveling at v0 = 20m/s hits a block of mass M = 450g. The block rests on
a frictionless horizontal surface and is attached to a spring of force constant k = 2000N/m. The projectile
ricochets backward off the block with speed v’ = 0.6v0. The collision lasts for 4ms.
(In the picture, the spring is attached to the block and the wall)
a) What impulse is delivered to the block and what is the average force for the collision?
b) Is the collision elastic? Justify your answer.
c) What is the maximum compression of the spring?
d) Suppose the surface has a “small” coefficient of friction μk = 0.1. Estimate the total distance the block
traverses from the moment it is struck to when it comes finally to rest. Why is this only an estimate? For
such an estimate to be accurate what is the criterion for “small” friction coefficient?

Homework Equations


p=mv
J=Δp
fav= Δp/Δt
KE=1/2mv2
Us=1/2kd2
(maybe?) vrel=-v'rel

The Attempt at a Solution


a)
momentum: mv0=mv'+Mv2'

impulse on block: J=Δp=p2-p1 --> p2= mv0 - mv' = .4mv0 = .4 N⋅s

fav= Δp/Δt = 100 N

b)
KE= 1/2 m v2
KE' +1/2mv'2+1/2Mv2'2

KE'/KE = [m(.6v0)+M(v0-.6v0)2] / mv02

=(.6)2 + (.450)(.4)2(.05) =>.40 --> 60% KE loss. non- elasticand this is where I am second guessing myself.
 
Last edited:
Physics news on Phys.org
.
 
Can't follow your working if you make up variables like p1, p2 etc. and don't define them.
Your answer to a) is wrong. Please explain your working in detail.
For part b), how are you calculating v2'?
 
x24759 said:
impulse on block: J=Δp=p2-p1 --> p2= mv0 - mv' = .4mv0 = .4 N⋅s

a) While the idea is not wrong here, you have forgotten to take into account that momentum is a vector quantity, which means that direction is important. Hence, it is important to define what direction you will be taking as positive and stick with that convention throughout your working. A diagram will help you here if you have trouble visualizing.

b) The problem that I mentioned above manifests in your working for part (b) as well. Look through the directions and signs of your vector quantities again.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top