Question

In Figure 9-69, block 1 of mass m1 slides from rest along a frictionless ramp from height h and then collides with stationary block 2, which has mass m2 = 3m1. After the collision, block 2 slides into a region where the coefficient of kinetic friction is μk 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? Express your answer in terms of the variables given and g.

Answer 1

part 1 : block 1 slides from high 2.5 m ,what velocity to collision block 2 ? use conservative energy
E2 =E1
PE1+EK1= PE2+Ek2
mgh+0 =0+(mv^2)/2
then v = sqrt (2gh) = sqrt (49) = 7 m/s

part 2 block 1 slide with velocity 7 to cllide block 2 after that what is velocity of block 2 ?
use conservation momentum
case 1- if elastic one dimention velocity block 2 is:
v2=[ 2 m1u1+u2(m2-m1)]/(m1+m2)
= (2 m1*7+0) /(m1+2m1)
= 2*7/3 =4.7 m/s
case 2 - if completely inelastic after collision become one body mass = m1+m2=3m1 and velocity is
v2= m1u1/(m1+m2)=m1*7/(m1+2m1)=7/3
= 2.33 m/s
part 3 slide on plane until stop v=0 use kinetic energy
KE2 - KE1 = total W
0 - 0.5mv^2 = - fk*d
- fk*d = work of kinetic friction
fk = uk mg =0.5*m1*g ,uk =coefficient of kinetic friction

case 1 elastic collision
0.5*2m1*4.7*4.7=0.5*2m1*9.8*d
9.8d = 4.7*4.7
d=2.254 m
case 2 Completely inelastic
0.5*3m1*2.33*2.33=0.5*3m1*9.8d
9.8d= 2.33*2.33
d = 0.539 m