In the figure below, block 1 of mass m 1 slides from rest along a frictionless r
ID: 1539787 • Letter: I
Question
In the figure below, block 1 of mass m1 slides from rest along a frictionless ramp from height h = 3.00 m and then collides with stationary block 2, which has mass m2 = 2.00m1. After the collision, block 2 slides into a region where the coefficient of kinetic friction k is 0.600 and comes to a stop in distance d within that region.
9. 0/1 points I Previous Answers HRW10 9.P.068 My Notes In the figure below, block 1 of mass m slides from rest along a frictionless ramp from height h 3.00 m and then collides with stationary block 2, which has mass m 2.00m After the collision, block 2 slides into a region where the coefficient of kinetic friction uk is 0.600 and comes to a stop in distance d within that region. Frictionless (a) What is the value of distance d if the collision is elastic? 2.66 X m (b) What is the value of distance d if the collision is completely inelastic? 667 X m Did you conserve mechanical energy to find the speed of block 1 at the bottom? Did you use one of the equations for a one-dimensional elastic collision to get the speed of block 2 just after the collision? Did you then calculate how far the block must slide for the kinetic frictional force to stop it? The procedure for the inelastic collision is similar except now only momentum is conserved in the collision.Explanation / Answer
v1=velocity of mass m1 before collision
v2=velocity of mass m2 before collision
v1’=velocity of mass m1 after collision
v2’=velocity of mass m2 after collision
v1=sqrt[2gh] = sqrt[2*9.8*3.0] = 7.67 m/s
v2=0m/s
a)
For elastic collision,
v2’ = 2m1v1/(m1+m2) = (2*m1*7.67)/(m1+2m1) = 5.11 m/s
For frictional surface we use law of conservation of energy,
KE = Wfriction = µk*m2*g*d
d= [1/2*m2*v2’^2]/[µk*m2*g] =[1/2*7.67^2]/[0.6*9.8] = 5.00m
b) For inelastic collision,
v2’ = m1v1/(m1+m2) = (m1*7.67)/(m1+2m1) = 2.56 m/s
For frictional surface we use law of conservation of energy,
KE = Wfriction = µk*m2*g*d
d= [1/2*m2*v2’^2]/[µk*m2*g] =[1/2*2.56^2]/[0.6*9.8] = 0.56m