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Question

A ball of mass 6 kg hits the floor with a speed 2 ms1 making an angle of incidence 30 with the normal. The coefficient of restitution of 13. Find the kinetic energy of ball just after reflection.

A
4 J
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B
48 J
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C
17 J
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D
13 J
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Solution

The correct option is A 4 J


During the collision, floor exert a normal force on the ball.

There is no external force parallel to the floor. Thus, the parallel component of the velocity of the ball remains unchanged.

2sin30=vsinθ

vsinθ=1 ......(1)

Using Newton's law of impact normal to the floor, we get,

e=velocity of separationvelocity of approach=v2v1u1u2

Here, (u1)y=2cos30 ; (u2)y=0 ms1 (v1)y=vcosθ ; (v2)y=0 ms1

13=vcosθ2×cos30

vcosθ=23cos30=13 ......(2)

From equation (1) and (2) we get,

v2sin2θ+v2cos2θ=(1)2+(13)2

v2=43

Now, the kinetic energy of the ball is,

K.E=12mv2

K.E=12×6×43=4 J

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (A) is the correct answer.
Why this question ?

Caution: In bouncing of a ball from the floor like questions, v=ev directly can be applied if the ball is colliding perpendicular to the floor. When this is not the case, this formula can be used for vertical component only.



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