A block is sliding down on an inclined plane of slope θ and at an instant t = 0 this block is given an upward momentum so that it starts moving up on the inclined surface with velocity u. The distance (S) travelled by the block before its velocity become zero, is ___

A block is sliding down on an inclined plane of slope θ and at an instant t = 0 this block is given an upward momentum so that it starts moving up on the inclined surface with velocity u. The distance (S) travelled by the block before its velocity become zero, is

Q. A block is sliding down on an inclined plane of slope $\theta$ and at an instant $t = 0$ this block is given an upward momentum so that it starts moving up on the inclined surface with velocity $u$. The distance $(S)$ travelled by the block before its velocity become zero, is ____ .

$(g = \text{gravitational acceleration})$

A. $\dfrac{2u^2}{g\cos\theta}$

B. $\dfrac{u^2}{\sqrt{2g\cos\theta}}$

C. $\dfrac{u^2}{2g\sin\theta}$

D. $\dfrac{u^2}{2g\cos\theta}$

Correct Answer: $\dfrac{u^2}{2g\sin\theta}$

Explanation

When the block moves upward on the inclined plane, the component of gravitational acceleration acting opposite to the motion is $g\sin\theta$.

The initial velocity of the block along the incline is $u$ and the final velocity becomes zero.

Using the kinematic equation:

$$ v^2 = u^2 - 2aS $$

Here, $$ v = 0,\quad a = g\sin\theta $$

Substituting, $$ 0 = u^2 - 2(g\sin\theta)S $$

Solving for $S$, $$ S = \frac{u^2}{2g\sin\theta} $$

Hence, the distance travelled by the block before coming to rest is $$ \boxed{\dfrac{u^2}{2g\sin\theta}} $$

Explanation

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