Question

A large, heavy box is sliding down a smooth plane of
inclination . From a point P on the bottom of the
box, a particle is projected inside the box. The initial
speed of the particle with respect to the box is a, and
the direction of projection makes an angle a with the
bottom as shown in figure. The distance along the
bottom of the box between the point of projection P
and the point where the particle lands is
(Assume that the particle does not hit any other
surface of the box. Neglect air resistance)​

Answer 1

Answer:

similar explanation in the photo also

Explanation:

Answer :

A::B::C

Solution :

(a)

is the relative velocity of the particle with respect to the box . Resolve

.<br>

x- direction

u_(y)

y- direction

y- direction

y

u_(y) = + u sin alpha

a_(y) = - g cos theta

s_(y) = 0 ( activity is taken till the time the particles comes back to the box .) <br>

<br>

<br>

<br>

- direction motion ( Taking relative terms w.r.t. box) <br>

<br>

<br> (b) For the observer ( on ground ) to see the horizontal displacement to be zero , the distance travelled by the box in time

should be equal to the range of the particle . <br> Let the speed of the box at the time of projection of particle be

. Then for the motion of box with respect to ground. <br>

<br>

<br>

<br> On solving we get