A ball moves on the horizontal surface of a billiards table with deceleration of constant magnitude
d m s−2
. The ball starts at A with speed 1.4 m s−1
and reaches the edge of the table at B, 1.2 s later,
with speed 1.1 m s−1
.
(i) Find the distance AB and the value of d. [3]
AB is at right angles to the edge of the table containing B. The table has a low wall along each of its
edges and the ball rebounds from the wall at B and moves directly towards A. The ball comes to rest
at C where the distance BC is 2 m.

Question

A ball moves on the horizontal surface of a billiards table with deceleration of constant magnitude 
d m s−2 
. The ball starts at A with speed 1.4 m s−1 
and reaches the edge of the table at B, 1.2 s later, 
with speed 1.1 m s−1 
. 
(i) Find the distance AB and the value of d. [3] 
AB is at right angles to the edge of the table containing B. The table has a low wall along each of its 
edges and the ball rebounds from the wall at B and moves directly towards A. The ball comes to rest 
at C where the distance BC is 2 m.

abmon98 · Answer

(ii) Find the speed with which the ball starts to move towards A and the time taken for the ball to 
travel from B to C. [3] 
(iii) Sketch a velocity-time graph for the motion of the ball, from the time the ball leaves A until it 
comes to rest at C, showing on the axes the values of the velocity and the time when the ball is 
at A, at B and at C

anonymous · Answer

(i)  $v_f=v_i+at$.         $v_f$ is the final velocity,  $v_i$ is the initial velocity, $a$ is the acceleration\deceleration, $t$ is the period of time.

$1.1=1.4+a\times 1.2 \rightarrow a = -0.25 m/s^2$


$v_f^2-v_i^2=2as$.     $s$ is the distance = AB
$\rightarrow 1.1^2-1.4^2=2\times -0.25 \times s \rightarrow s =1.5 m$

anonymous · Answer

(ii)  Is the deceleration when the ball moves from B to C $-0.25m/s^2$?

abmon98 · Answer

i solved i) the problem is with understanding ii) a bit

anonymous · Answer

Great. Just a little effort and we can do everything.

abmon98 · Answer

i want to know why the acceleration to approach the wall is equal when the ball rebounds

anonymous · Answer

I don't quite sure. I just searched for the answer but couldn't find an appropriate one for this.

abmon98 · Answer

i solved the question

abmon98 · Answer

but i dont understand why the acceleration before is the same acceleration when rebounded.

abmon98 · Answer

the deceleration to approach B is the same deceleration after re bouncing. -1/2(-.25)t^2=2 0.125t^2=2 t=4seconds a=v-u/t-->-0.25*4=0-u u=1m/s

anonymous · Answer

Yes, you are right.

anonymous · Answer

I had the same question when I studied about this. Why the acceleration is the same???

abmon98 · Answer

@ProfBrainstorm