Question

How do you solve this physics problem ?

there are two A and B balls on a pool table at rest, one next to the other. After the impulse, the A-ball moves with an
acceleration of 12 m/s^2 and the B ball with an acceleration of 24 m/s^2. If the angle formed between the two balls is 60°, what will be the distance, in cm, between the two balls after one second, considering that neither of them has fallen into the hole?

THANK YOU

Answer 1

First we find the distance covered by each ball, then the distance between them. They are `2080` `cm` apart after 1 second.

Explanation:

For both balls, the initial velocity `u=0` `ms^-1`, so:

`s=ut+1/2at^2`

reduces to:

`s=1/2at^2`

For Ball A:

`s=1/2*12*1^2=6` `m`

For Ball B:

`s=1/2*24*1^2=12` `m`

We can't simply subtract one from the other, though, because they are at a `60^o` angle to each other. We have a triangle with sides `6` and `12` `m` and a `60^o` angle between them. We can use the cosine law to find the third side of the triangle:

`a^2=b^2+c^2-2bc cosA`

`=12^2+24^2-2*12*24*cos60^o`

`=144+576-576*0.5=432`

Take the square root of both sides:

`a=20.8` `m`

We were asked for the answer in `cm`, so it is `2080` `cm`.

(side note: this is clearly not a standard size pool or billiard table!)