Question

A 2kg ball moving at 2 m/s is rolling towards an inclined plane. It eventually rolls up the hill to a position near the top of the ramp where it momentarily stops before it rolls back down. How high up the ramp does the ball roll?

Answer 1

To find the height up the ramp a 2kg ball will roll before stopping, we use the conservation of energy. The ball's initial kinetic energy, which is 4 J, will equal its gravitational potential energy at the top. Thus, the ball rolls up to a height of 0.2 meters.

Step-by-step explanation:

The student's question involves determining how high up an inclined plane a 2kg ball that is moving at 2 m/s will roll before it momentarily stops. This is a physics problem that can be solved using the principle of conservation of energy. The kinetic energy of the ball at the bottom of the ramp is converted into potential energy at the top, where it momentarily stops.

The initial kinetic energy (KE) of the ball is given by the formula:

KE = \(\frac{1}{2}m v^2\)

Where 'm' is the mass of the ball and 'v' is the velocity. For our scenario with 'm' = 2kg and 'v' = 2 m/s, we have:

KE = \(\frac{1}{2} \times 2kg \times (2 m/s)^2 = 4 J\)

This kinetic energy is completely converted into gravitational potential energy (PE) at the height 'h' which can be expressed as:

PE = m g h

With 'g' being the acceleration due to gravity (10 m/s²) and 'h' representing the height in meters. To find 'h', we set PE equal to KE :

4 J = 2 kg \times 10 m/s² \times h

h = \(\frac{4 J}{20 kg \cdot m/s^2}\)

h = 0.2 m

Therefore, the ball will roll up the ramp to a height of 0.2 meters before it stops and rolls back down.

Answer 2

Remember that energy is always conserved, so all the kinetic energy of the ball will be converted to potential energy at the top of its roll.
1/2 (2) (2)^2 = 2*9.81*h
h= 0.204 m g is taken 9.81 ms-1