Question

In a head-on elastic collision of two bodies of equal masses
(a) the velocities are interchanged
(b) the speeds are interchanged
(c) the momenta are interchanged
(d) the faster body slows down and the slower body speeds up.

Answer 1

Answer:

d the faster body slows down and the slower body speeds up

Answer 2

(a), (b), (c) and (d) are correct  

Explanation:

Step 1:

If u and v are the pre-collision velocities and u ' and v ' are the post-collision velocities, then

$u^{\prime}=\frac{(\mathrm{m}-\mathrm{m}) \mathrm{u}}{\mathrm{m}+\mathrm{m}}+\frac{2 \mathrm{mv}}{(\mathrm{m}+\mathrm{m})}$

$u^{\prime}=\frac{(0) \mathfrak{u}}{2 \mathrm{m}}+\frac{2 \mathrm{mv}}{(2 \mathrm{m})}$

$=0+v=v$

and  

Step 2:

$\mathrm{v}^{\prime}=\frac{2 \mathrm{m} \mathrm{u}}{\mathrm{m}+\mathrm{m}}+\frac{(\mathrm{m}+\mathrm{m})}{(\mathrm{m}+\mathrm{m})}$

$\mathrm{v}^{\prime}=\frac{2 \mathrm{mu}}{2 \mathrm{m}}+\frac{(0) \mathrm{v}}{(2 \mathrm{m})}$

$=u+0=u$

So the velocities and speeds are interchanged. That is why (a) and (b) are true.

Since the velocities are changed and the masses are the same the moments are interchanged as well.  

Therefore (c) it is true.

If $u>v$, the speeds of the bodies will be changed after the collision.Now the body slows faster and the body speeds up gradually. So (d) it's true.