How can collisions and energy be related in a horizontal table scenario?

  • Thread starter Thread starter JJBrian
  • Start date Start date
  • Tags Tags
    Collisions Energy
AI Thread Summary
The discussion focuses on a physics problem involving a 0.1 kg mass colliding with a 1.2 kg spring gun on a horizontal table. Key calculations include determining the recoil speed of the spring gun after maximum compression, which involves using energy conservation principles. The energy stored in the spring is calculated to be 1.8 J, while the maximum compression of the spring is found to be 3 m. Additionally, if the mass collides with a block of putty instead, the recoil speed is calculated to be approximately 0.4615 m/s. The discussion emphasizes the importance of following a systematic approach to solving the problem.
JJBrian
Messages
18
Reaction score
0

Homework Statement


A 0.1kg is shot with a speed of 6m/s toward a 1.2kg spring gun( with spring constant of 0.4N/m). The spring gun is initially at rest with its spring relaxed. The spring gun is free to slide without friction on a horizontal table. The 0.1 kg mass compresses the spring to its maximum and remains lodged at this maximum compression.

a)what is the recoil speed of the spring gun( with the 0.1kg mass) after this event?

b)What is the energy stored in the spring gun after this event?

c) How much is the spring compressed from its relaxed position?

d) If instead of hitting a spring gun, this 0.1kg mass hit a 1.2 block of putty ( and stuck to the putty) that was free to slide with no friction on a horizontal table, what would be the recoil speed of the putty( with the 0.1 kg mass)?
 
Physics news on Phys.org
Attempt ... I think my approach is totally wrong
I started with part c first...
I think the question should be answer in order.
The problem is I don't know how to approach question a with the given variables.
a)
Ws = (1/2)kx^2max
Ws = (0.4N/m)(3m)^2
Ws = 1.8J
Ws = 1/2mvf^2-1/2mvi^2
vf=sqrt(vi^2 +(2/m)*Ws
vf=sqrt(6m/s^2 +(2/.1kg)*(1.8J)
b)
Us = 1/2kx^2
Us = 1/2(0.4N/m)(3m)
Us = 0.6J
c)KE + Us = KE+ Us
0 +1/2kx^2max = 1/2mv^2 + 0
xmax =sqrt(m/k)*V
xmax = sqrt(.1kg/.4N/m)*(6m/s)
xmax = 3m
d) Vf = (m1/m1+m2)*v0
Vf = (.1kg/.1kg+1,2kg)*(6m/s)
Vf = 0.4615m/s
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top