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A ball is projected from the ground with speed u at an angle α with horizontal. It collides with a wall at a distance ‘a’ from the point of projection and returns to it original position. Find the coefficient of restitution between the ball and the wall.

ASh , 7 Years ago
Grade 11
anser 1 Answers
venkat
We know that,
Coefficient of restitution e=(v2-v1)/(u1-u2)
Initial velocity of the ball u1=ucos\alpha\vec{i}+usin\alpha \vec{j}
Initial velocity of the wall u2=\vec{0}
Final velocity of the ball v1=ucos(\pi+\alpha)\vec{i}+usin(\pi+\alpha)\vec{j}= -ucos\alpha \vec{i}-usin\alpha \vec{j}
Final velocity of the ball v2=\vec{0}
Now,
e={(\vec{0})-( -ucos\alpha \vec{i}-usin\alpha \vec{j})}/{(ucos\alpha\vec{i}+usin\alpha \vec{j})-(\vec{0})}
e=(ucos\alpha\vec{i}+usin\alpha \vec{j})/(ucos\alpha\vec{i}+usin\alpha \vec{j})
\therefore e=1
Last Activity: 7 Years ago
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