The collision between two pucks

  • Thread starter Thread starter agadag
  • Start date Start date
  • Tags Tags
    Collision
AI Thread Summary
The discussion revolves around a physics problem involving a collision between two pucks with different masses and initial velocities. The user initially struggles with setting up and solving the equations for momentum conservation in both the x and y directions. After some back-and-forth, a correct solution is provided, yielding final speeds of Puck A at 3.41 m/s and Puck B at 2.57 m/s. The importance of careful arithmetic and substitution in solving the equations is emphasized, highlighting a common pitfall in such problems. The exchange concludes with appreciation for the assistance received in clarifying the solution process.
agadag
Messages
59
Reaction score
0

Homework Statement


Puck A has a mass of 0.294 kg and is moving along the x-axis with a velocity of 5.55 m/s. It makes a collision with puck B, which has a mass of 0.588 kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with the angles shown in the drawing. Find the final speed of
Puck A
Puck B

Homework Equations


I have spent an hour on this problem. I have two different eqns one for x and one for y, respectively:
m1v0 = m1v1 cos 65 + m2v2cos 37
and
0 = m1v1 sin 65 - m2v2 sin 37



The Attempt at a Solution


I solved for v2 from the eqn for y and tried to plug that value into the eqn for x. This did not work and I am running out of ideas and patience. Please help! There is also a pic for this problem..
 

Attachments

  • physics.jpg
    physics.jpg
    21.5 KB · Views: 839
Physics news on Phys.org
I think you accidentally messed up you arithmetic when solving the question. Your X and Y equations are correct so you've set up your question correctly. i think you have just made a mistake when solving because the substitution of V2' of you Y equation into your X equation is the correct thing to do.

I got V1' = 3.41 m/s; V2' = 2.57 m/s; And these numbers make sense in my mind
 
Your numbers worked out perfectly. I don't understand where i am going wrong, because i am not getting your numbers. can you do a step by step please? id really appreciate it.
 
X-component: m1v1 = m1v1' cos 65 + m2v2'cos 37
Y-Component: 0 = m1v1' sin 65 - m2v2' sin 37

Y: 0 = m1v1' sin 65 - m2v2' sin 37
m1v1' sin 65 = m2v2' sin 37
m1v1' sin65/(m2sin37) = v2' -> Plug in numbers
(0.294 kg)*(v1')*(sin65)/[(0.588 kg)(sin37)] = v2'
0.75298*v1' = v2' -> substitute into the X

m1v1 = m1v1' cos 65 + m2v2'cos 37
(0.294 kg)(5.55 m/s) = (0.294 kg)v1'(cos65) + (0.588 kg)(0.75298v1')(cos37)
1.6317 kg*m/s = 0.12425*v1' kg + 0.3536*v1' kg
1.6317 kg*m/s = 0.4778*v1' kg
v1' = 3.41 m/s -> substiture back into Y

0.75298*v1' = v2'
0.75298*(3.41 m/s) = v2'
v2' = 2.57 m/s

So this was my solution... a lot messier on the computer but i hope you can find where you and i differed in our solutions and this can help you avoid the same mistake in the future.

Cheers
 
Thankyou so much! This was such a help to say the least
 
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...
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}...
Back
Top