Mass of block A is 3 kg and it begins its journey at the top of a frictionless i
ID: 1485416 • Letter: M
Question
Mass of block A is 3 kg and it begins its journey at the top of a frictionless incline with a velocity of 5 m/s down the incline. The patch following the incline is also frictionless. Mass of block B is 2 kg and it is sitting at the bottom of the incline as shown in the figure below.
The two blocks undergo a perfectly inelastic collision. Assuming the two blocks to be point masses and neglecting air-resistance, calculate the following:
(a) the pre-collision work done by gravity on block A
(b) the pre-collision speed of block A
(c) the post-collision speed of the two blocks
Explanation / Answer
a)
work done by gravity = mA*g*h
= 3*9.8*3
= 88.2 J
b)
use conservation of energy for block A:
kinetic energy at top + work done by gravity = kinetic energy at bottom
0.5*mA*VAi^2 + 88.2 = 0.5*mA*VAf^2
0.5*3*5^2 + 88.2 = 0.5*3*VAf^2
125.7 = 0.5*3*VAf^2
VAf = 9.15 m/s
Answer: 9.15 m/s
c)
use conservation of momentum:
mA*VA= (mA + mB) * Vf
3*9.15 = (3+2)*Vf
Vf = 5.5 m/s
Answer: 5.5 m/s
d)
Lost in kinetic energy = kinetic energy before collision - kinetic energy after collision
= 0.5*mA*VA^2 - 0.5*(mA+mB)*Vf^2
= 0.5*3*9.15^2 - 0.5*(3+2)*5.5^2
= 50 J
Answer: 50 J