A small block of mass M=1kg moves on a frictionless surface on an inclined plane. The surface of inclined PQ changes from 60° to 30° at Q. The include of QR is 30°. The block is initially at rest at P. Assuming that there is no loss of energy for collision between the block and the incline(g=10m per s^2),
Calculate the speed of block at Q immediately before going on second inclined and the energy at R immediately as it leaves the plane of 30° inclination.
Answers
Answer:
According to the question, let’s illustrate it using a figure as below:
Between the points P and Q, the height that the block falls is
h1 = √3 tan60° = 3m
∴ The Speed of the block just before striking the second incline,
v1 = √2gh1= √2*10*3 = √60 m per s^1
In a perfectly inelastic collision, component of v1 perpendicular to QR will become zero, while component of v1 parallel to QR will remain unchanged
∴ The Speed of block Q immediately after it strikes the second inline is,
v2 = component of v1 along QR
= v1 cos 30° = (√60) (√3/2)
= √45 m per s^1
The Height that the block falls from Q to R:
h2 = 3√3 tan 30° = 3 m
Let v3 be the speed of the block, at point R, just before it leaves the second incline, then:
v3 = √v2(square)+ 2gh2
= √(45+2*10*3)
= √105 m per s^1
