Question

If the objects have masses m1=1.19 kg and m2=4.48 kg, Object 1 is initially moving with a velocity v=7.98 m/s while object 2 is at rest.

What are the final velocities of the objects after the collision if it is perfectly elastic?

Answer 1

nearly every time a problem says, 'collision' its really saying use the law of momentum.
momentum, p, is equal to mass times velocity. p = m*v

in collision problems it is important to break them down into two parts; the first is the part where we look at the mass and velocities of all objects before the collision. and the second part is to look at the mass and velocities again but after the collision. (for now i am skipping the concept of elasticity, because it says the collision is perfectly elastic. perfectly elastic is a smart way of saying, 'no momentum is lost through friction or heat at the moment the objects collide.'


event 1 = event 2

or can be said as..
total momentum of before = total momentum after collision

or stated as...
(momentum of object 1 + momentum of object 2) before collision = (momentum of object 1 + momentum of object 2) after collision

and the math version of the two phrases above is like this:
mv + mv = mv + mv

so the 'before collision' side of the equation will look like this..
1.19 kg (7.98 m/s) + 4.48 kg( 0 m/s) =

the problem is worded so that we are to understand that the objects will have the same velocity after the collision. so the 'after collision' side of the equation will go from looking like this:
= mv + mv
to
= v(m + m) which is saying to add the mass of object 1 and object 2, then multiply it by the velocity. this velocity is the 'final velocity' in which we are solving for.

so using the 'before collision' side of the equation previously discussed and plugging in mass 1 and mass 2 into the 'after collision' side of the equation and with some algebra, you can solve for v, final velocity.

Answer 2

I can't give you the answer, but it should be between 0 and 3 m/s

Answer 3

Wouldn't that solution be only for an inelastic situation m1v1+m1v2 = v(m2+m1). So it wouldn't apply towards this one. Wouldn't it be more along the lines of
\[v_{1f}= ((m_{1}-m_{2})/(m_{1}+m_{2})) V _{1} i + 2m _{2}\]
\[v_{2f}= (2m_{1}/m_{1}+m_{2})V_{1}i + (m_{2}-m_{1})/(m_{1}+m_{2}) V_{2}i\]