In the figure, block 1 of mass m 1 slides from rest along a frictionless ramp fr
ID: 1325296 • Letter: I
Question
In the figure, block 1 of mass m1 slides from rest along a frictionless ramp from height h = 2 m and then collides with stationary block 2, which has mass m2 = 5m1. 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?
Can you show the steps taken?
In the figure, block 1 of mass m1 slides from rest along a frictionless ramp from height h = 2 m and then collides with stationary block 2, which has mass m2 = 5m1. 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? Can you show the steps taken?Explanation / Answer
a) first to compute the velocity of m1 just before collision ,
using energy conservation,
mgh = mu^2 /2
u =sqrt(2gh) = sqrt(2*9.81*2) = 6.26 m/s
now using momentum conservation,
m1 * 6.26 + m2*0 = m1*(-v1) + m2*v2
and m2 = 5m1
6.26m1 = -m1v1 + 5m1v2
5v2 - v1 = 6.26 ..... (i)
for elastic collision,
velocity of sepration = velocity of approach
v1 +v2 = 6.26 .... (ii)
solving (i) and (ii) ,
v1 = 4.17 m/s
v2 = 2.09 m/s
in horizontal ,
Fnet = friction force = ma
u.mg = ma
a =u.g = 0.6 x 9.81 = 5.89 m/s2
using v^2 - u^2 = 2ad
0 - 2.09^2 = 2 x -5.89 x d
d = 0.371 m
b) inelastic collision ,
first to compute the velocity of m1 just before collision ,
using energy conservation,
mgh = mu^2 /2
u =sqrt(2gh) = sqrt(2*9.81*2) = 6.26 m/s
now using momentum conservation,
m1 * 6.26 + m2*0 = (m1 + m2)v
and m2 = 5m1
6.26m1 = (m1 + 5m1)v
v =1.04 m/s
in horizontal ,
Fnet = friction force = ma
u.mg = ma
a =u.g = 0.6 x 9.81 = 5.89 m/s2
using v^2 - u^2 = 2ad
0 - 1.04^2 = 2 x -5.89 x d
d = 0.0924 m