The angle that the stuck ball makes with the direction of motion and the velocity of both the billiard balls are 65°, 2.6 m/s and 5.5m/s respectively.
To find the answer, we need to know about the linear momentum and angle made in elastic 2 dimensional collision.
What's the angle made by the struck billiard ball with the direction of motion of the 1st ball?
- In the two dimensional elastic collision between two objects, the scattered objects make 90° angle between them.
- So, here the angle between the two billiard balls after collision is 90°.
- As the one billiard ball is making 25° with the initial direction of motion, so the another struck billiard ball makes 90°-25°= 65°.
What's the conservation of linear momentum along vertical direction?
- Let m = mass of each billiard balls, v = velocity of scattered billiard ball, u = velocity of struck billiard ball
- Before collision there's no vertical component of momentum. So it's zero.
- From figure, the total momentum along vertical direction is m×v×sin25°-m×u×sin65°
- From conservation of linear momentum,
0= m×v×sin25°-m×u×sin65°
v×sin25°=u×sin65°
v= 2.1u
What's the conservation of momentum along horizontal direction as shown in figure?
- Total momentum before the collision along horizontal direction = m × 6 m/s
- From the figure, total momentum after collision is m×v×cos25° + m×u×cos65°
- So, m×6 = m×v×cos25° + m×u×cos65°
6= 2.1u× cos25° +u×cos65°
= 2.33u
- u = 6/2.33 = 2.6 m/s
- And v= 2.1×2.6= 5.5 m/s
Thus, we can conclude that the angle that the stuck ball makes with the direction of motion and the velocity of both the billiard balls are 65°, 2.6 m/s and 5.5m/s respectively.
Learn more about the 2 dimensional collision here:
brainly.com/question/14445985
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