In the figure, block 1 of mass m 1 slides from rest along a frictionless ramp fr
ID: 1282328 • Letter: I
Question
In the figure, block 1 of mass m1 slides from rest along a frictionless ramp from height h = 2.5 m and then collides with stationary block 2, which has mass m2 = 2m1. After the collision, block 2 slides into a region where the coefficient of kinetic friction ?k is 0.6 and comes to a stop in distance d within that region. What is the value of distance d if the collision is (a) elastic and (b) completely inelastic?
In the figure, block 1 of mass m1 slides from rest along a frictionless ramp from height h = 2.5 m and then collides with stationary block 2, which has mass m2 = 2m1. After the collision, block 2 slides into a region where the coefficient of kinetic friction ?k is 0.6 and comes to a stop in distance d within that region. What is the value of distance d if the collision is (a) elastic and (b) completely inelastic?Explanation / Answer
The speed of block 1 before impact
m*g*h = 1/2*m*v^2 so v = sqrt(2*g*h) = sqrt(2*9.80*2.5) = 7m/s
a) elastic (conservation of momentum) m1*7 = m1*v1 + 3m1*v2
or 7= v1 + 3*v2 where v2 is the speed of m2 after the collision
so v1 = 7 - 3v2
& K is conserved also so 1/2*m1*7^2 = 1/2*m1*v1^2 + 1/2*3*m1*v2^2
or 49 = v1^2 + 3v2^2
now 49 = (7 - 3v2)^2 + 3v2^2
or 12v2^2 - 42v2 = 0
so v2 = 42/12 = 3.5m/s
Now ?_k*3m*g*d = 1/2*3m*v2^2
so d = 1/2*3^2/(9.8*?_k) = 1/2*3^2/9.8*0.6) = 0.7653m
b) the collision is completely inelastic so v2 = m1*7/(m1+2m1) = 2.33m/s
Now ?_k*4m*g*d = 1/2*4m*v2^2
so d = 1/2*2.33^2/(9.8*?_k) = 1/2*2.33^2/9.8*0.6) = 0.46296m