Question

# A ball of mass 6 kg hits the floor with a speed 2 ms−1 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=v2−v1u1−u2 Here, (u1)y=2cos30∘ ; (u2)y=0 ms−1 (v1)y=vcosθ ; (v2)y=0 ms−1 ⇒13=vcosθ2×cos30∘ ⇒vcosθ=23cos30∘=1√3 ......(2) From equation (1) and (2) we get, v2sin2θ+v2cos2θ=(1)2+(1√3)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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