Elastic Collision: 1kg Object, 0.7kg Object, 0.2s Interval

[ironhawk316]

Homework Statement

A 1 Kilogram object with speed of 2.0 m/s in the positive x direction has a head -on elastic collision with a stationary 0.70Kg object located at x=0. What acceleration does the stationary object experiences during the collision if the collision occurs over 0.2 seconds?

Homework Equations

m1v1i + m2v2i = m1v1f + m2v2f

The Attempt at a Solution

m1v1i + m2v2i = m1v1f + m2v2f
(1)(2)+(0.7)(0)=(1)(v1f)+(0.7)(v2f)
v1 +0.7 v2f = 2

plug in:

v1+0.7(2+v1)=2
V1+ 1m +0.7 v1=2
1.7=2-1.4
1.7 v1=0.6
v1=0.6/1.7 =0.353 m/s

v2=2+0.353= 2.353 m/s

m2(v2i-v2f)
=0.7(2.353 -0)
=1.6471=ft

F=1.6471/t=1.6471/0.2=8.24N

a=F/m = 8.24/1= 8.24 m/s^2 final answer

Is this correct? Where did I go wrong? Do I have the everything in the right place?

 

[rl.bhat]

Using conservation of momentum and energy find v2f. The rate of change of velocity will give you the acceleration of the stationary body.

 

[nlightner]

Your solution looks correct. However, I would like to clarify that the acceleration experienced by the stationary object during the collision would be equal to the change in velocity divided by the time interval, rather than just the final velocity. So the correct equation would be:

a = (v2f - v2i) / t

In this case, the final velocity of the stationary object would be 2.353 m/s and the initial velocity would be 0 m/s, so the acceleration would be:

a = (2.353 - 0) / 0.2 = 11.765 m/s^2

Other than that, your calculations and understanding of the problem seem to be correct. Good job!