Two masses sit at the top of two frictionless inclined planes that have different angles. Which mass gets to the bottom first?

both mass take equal time to reach at bottom

Explanation:

As both masses M 1 and M 2 are situated at same height so that the potential energy at top is transformed to kinetic energy at the bottom by applying principle of energy conservation.

PE = KE

Mgh = (1/2) (M) (V^2)

V^2 = 2gh

V = (2gh) ^ (1/2)

So velocity is independent of mass

For M1

(M_1) (g) (h) = (1/2) (M1) (V^2)

V^2 = 2gh

V = (2gh) ^ (1/2)

For M2

(M_2) (g) (h) = (1/2) (M2) (V^2)

V^2 = 2gh

V = (2gh) ^ (1/2)

therefoer speed at bottom too are independent of mass

So we can say both speed are independent of therir respective mass

hence both take equal time to reach at bottom

Let the two inclined planes having angle of inclinations α and β.

The acceleration along the inclined plane acting on the body is gSinα and gSinβ.

If α > β

So, g Sinα > g Sinβ

So, more the inclination of the plane more be the acceleration of body and hence the time taken is less.

So, the body kept on the inclined whose inclination is more reaches at the bottom first.