skolepikegq9s

2023-02-21

Two billiard balls undergo a head-on collision. Ball 1 is twice as heavy as ball 2. Initially, ball 1 moves with a speed u towards ball 2 which is at rest. Immediately after the collision, ball 1 travels at a speed of $\frac{u}{3}$ in the same direction. What type of collision has occurred?

A)Inelastic;

B)Elastic;

C)Completely inelastic;

D)cannot be determined from the information

A)Inelastic;

B)Elastic;

C)Completely inelastic;

D)cannot be determined from the information

Kamari Potts

Beginner2023-02-22Added 10 answers

The correct answer is B Elastic

Let mass of ball 1=2m

Mass of ball 2=m

Velocity of ball 1 before collision =u

Velocity of ball 1 after collision $=\frac{u}{3}$

Velocity of ball 2 before collision =0

Let final velocity of ball 2$={v}_{2}$

Momentum will be conserved because there is no outside force acting on the system.

Prior to and following a collision, using conservation of linear momentum.

$(2m)u+m(0)=(m){v}_{2}+(2m)(\frac{u}{3})$

or ${v}_{2}=\frac{4u}{3}$

Now,

$e=\frac{\text{velocity of separation}}{\text{velocity of approach}}\phantom{\rule{0ex}{0ex}}e=\frac{{v}_{2}-{v}_{1}}{{u}_{1}-{u}_{2}}\phantom{\rule{0ex}{0ex}}e=\frac{\frac{4u}{3}-\frac{u}{3}}{u-0}=1$

Thus, collision is elastic.

Let mass of ball 1=2m

Mass of ball 2=m

Velocity of ball 1 before collision =u

Velocity of ball 1 after collision $=\frac{u}{3}$

Velocity of ball 2 before collision =0

Let final velocity of ball 2$={v}_{2}$

Momentum will be conserved because there is no outside force acting on the system.

Prior to and following a collision, using conservation of linear momentum.

$(2m)u+m(0)=(m){v}_{2}+(2m)(\frac{u}{3})$

or ${v}_{2}=\frac{4u}{3}$

Now,

$e=\frac{\text{velocity of separation}}{\text{velocity of approach}}\phantom{\rule{0ex}{0ex}}e=\frac{{v}_{2}-{v}_{1}}{{u}_{1}-{u}_{2}}\phantom{\rule{0ex}{0ex}}e=\frac{\frac{4u}{3}-\frac{u}{3}}{u-0}=1$

Thus, collision is elastic.