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.