Answer:
Explanation:
We denote by u the velocity of the prism (Fig. 214), by vx and vy the horizontal and vertical components of the velocity v of the body relative to the ground, and by β the angle between
tanβ = vy/vz. (1)
Since there acts on the prism in the vertical direction, apart from the force from the body, a force due to the reaction from the support, then, in considering the behaviour of the system of the body and the prism, we may apply the law of conservation of momentum only to the horizontal component of the velocities of the body and the prism. Thus the velocities u and v0 satisfy the relation
Mu = mvx . (2)
We assume that at some instant the body is at point A on the prism (Fig. 215). During the next second, the prism will move u cm to the left, the body will move vx cm to the right in the horizontal direction, vx and vy cm in the vertical direction. The magnitudes
of these displacements should be such that the body will be at some point B. Consequently, velocities vx and vy should satisfy not only the laws of conservation of energy and momentum, but also the relation
This relation is an expression of the condition that the body always remains in contact with the prism during the motion. From relation (2) we find
u = (m/M)vx.
Inserting the value of u into equation (3), making use of equation (1), and rearranging, we obtain
As should have been expected tanβ > tanα, and β > α.
The velocity of a body sliding down a moving prism is directed at a larger angle to the horizontal than in the case of a stationary prism. Using the law of conservation of energy and knowing the height of the initial position of the body, we can calculate the magnitudes of the velocities u and v.