Question

A cart is sliding down a low friction incline. A device on the cart launches a ball, forcing the ball perpendicular to the incline, as shown above . Air resistance is negligible. Where will the ball land relative to the cart, and why?

• A. The ball will land in front of the cart, because the balls acceleration component parallel to the plane is greater than the carts acceleration component parallel to the plane
• B. The ball will land in front of the cart, because the ball has a greater magnitude of acceleration than the cart
• C. The ball will land in the cart, because both the ball and the cart have the component of acceleration parallel to the plane
• D. The ball will land in the cart, because both the ball and the cart have the same magnitude of acceleration

Answer 1

The correct option is C The ball will land in the cart, because both the ball and the cart have the component of acceleration parallel to the plane

Let the angle of inclination be $\theta$

Force due to gravity acting on cart perpendicular to plane is:

$W_{1N}=Mg\cos \theta$ which is balanced by the normal reaction. Hence, the cart has no displacement perpendicular to the plane.

Force due to gravity acting on cart parallel to plane is:

$W_{1T}=Mg\sin \theta$

Acceleration of the cart parallel to the plane is $a_{1T}=g\sin \theta$

Similarly, acceleration of the ball parallel to the plane is $a_{2T}=g\sin \theta$

Relative acceleration of the ball wrt the cart parallel to the plane is:

$a_r=a_{2T}-a_{1T}=0$

Initial relative velocity of the ball wrt the cart parallel to the plane is also zero as it is projected perpendicular to the plane.

Hence, relative displacement of the ball wrt the cart parallel to the plane remains zero at all times. Thus, the ball lands back in the cart.