Mass of block A is 3 kg and it begins its journey at the top of a frictionless i

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 inclastic collision. Assuming the two blocks to be point masses and neglecting air-resistance, calculate the following: The pre-collision work done by gravity on block A the pre-collision speed of block A the post-collision speed of the two blocks the total mechanical energy lost during the collision

Explanation / Answer

a)
work done by graviy = F*h
= m*g*h
=3*9.8*3
= 88.2 J
Answer: 88.2 J

b)
use work energy theorem
Kinetic energy at top + work done by gravity = kinetic energy at bottom
0.5*m*Vi^2 + 88.2 = 0.5*m*Vf^2
0.5*3*5^2 + 88.2 = 0.5*3*Vf^2
Vf = 9.15 m/s
Answer: 9.15 m/s

c)
use conservation of momentum while collsion
Ma*Va = (Ma+Mb) * V
3*9.15 = (3+2)*V
V= 5.5 m/s
Answer: 5.5 m/s

d)
mechnical energy lost = kinetic energy before collision - kinetic energy after collision
= 0.5*(3)*(9.15)^2 - 0.5*(3+2)*(5.5)^2
= 50 J
Answer: 50 J

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