Question

Two objects of different mass slide down two different ramps without friction. Both objects are released from rest from the same height, but one ramp is much steeper than the other. m1 > m2.

What are the relative speeds of the two masses when they reach the bottom?

Answer 1

They have the same speed

Step-by-step explanation:

Law of Conservation of Mechanical Energy

In the absence of dissipative forces (like friction, air resistance), the total amount of mechanical energy, in a closed system remains constant. The mechanical energy is the sum of the potential gravitational and kinetic energies:

`M=K+U`

`\displaystyle M=(1)/(2)mv^2+mgh`

Where m is the mass of the object, v its speed and h its height.

The first object's mechanical energy just after it's released in the ramp is

`\displaystyle M_(1)=(1)/(2)m_1v_1^2+m_1gh`

Since it's initially at rest

`M_(1)=m_1gh`

When it reaches the bottom of the ramp, all it mechanical energy becomes kinetic, so

`\displaystyle m_1gh=(1)/(2)m_1v_1'^2`

Being v1' the final speed at the bottom of the ramp. Solving for v1'

`v_1'=√(2gh)`

The second object's mechanical energy just after it's released in the ramp is

`\displaystyle M_(2)=(1)/(2)m_2v_2^2+m_2gh`

Note the height is the same for both objects. Following the same procedure with m2, we get

`v_2'=√(2gh)`

Or, similarly

`v_1'=v_2'`

We can see both speeds are the same regardless of their masses or the steepness of the ramps they came from