A small block of mass m 1 = 0.500 kg is released from rest at the top of a curve

Question

A small block of mass m1 = 0.500 kg is released from rest at the top of a curved-shaped, frictionless wedge of mass m2 = 3.00 kg, which sits on a frictionless horizontal surface as in the figure below. When the block leaves the wedge, its velocity is measured to be v1 = 4.60 m/s to the right.

(a) What is the velocity of the wedge after the block reaches the horizontal surface?
... to the left

(b) What is the height h of the wedge?

...

I want precisely-calculated answers with units included.

Explanation / Answer

(a) By Conservation of momentum

When the block reaches the horizontal surface its momentum is
0.5*4.60= 2. 3 kg.m/s

The wedge will have the same momentum in the opposite direction
= 3.00 * v = 2.3 kg.m/s
v = 2.3/ 3 = 0.767 m/s

The velocity of the wedge to the left is 0.767 m/s.

(b) The Kinetic Energy of the block
= 0.5 * 0.5 * 4.69^2 = 5.5 J

KE of the wedge
= 0.5 * 3.0 * 0.767^2 =0.8824 J

Total kinetic energy = 5.5 + 0.8824 = 6.3824 J

This equals the potential energy lost by the block
= mgh = 0.5 * 9.81 * h = 6.3824
h = 6.3824 / (0.5 * 9.81) = 1.301 m

The height of the wedge is 1.444 m.

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