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A brick is projected from ground with speed v at angle `theta` from horizontal. The longer face of brick is parallel to ground. The brick slides along ground through some distance after hitting ground and then stops. The collision is perfectly inelastic. The coefficient of friction is `mu(mu lt cot theta)`. The angle `theta` is chosen such that brick travels the maximum horizontal distance before coming to rest. Find the distance in meters : |
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Answer» Correct Answer - A::B Range `= (v^(2) sin 2 theta)/(g)` Impulse along vertical after collision `I_(x) = mv sin theta` Horizontal impulse `I_(y) = - mumv sin theta` So horizontal velocity after impact `= v cos theta - mi v sin theta` So total distance along horizontal `D=(v^(2)2 sin cos theta)/(g)+(v^(2))/(2mu g)(sin theta - mu cos theta)^(2)` `=(v^(2)(4 mu sin theta cos theta+(sin theta-mu cos theta)^(2)))/(2 mu g)` `=(v^(2)(sin theta + mu cos theta)^(2))/(2mug)` `D_(max) = (v^(2)(1+mu^(2)))/(2mu g)`. |
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