Mass m = 0.1 kg moves to the right with speed v = 0.29 m/s and collides with an
ID: 1613334 • Letter: M
Question
Mass m = 0.1 kg moves to the right with speed v = 0.29 m/s and collides with an equal mass initially at rest. After this inelastic collision the system retains a fraction = 0.8 of its original kinetic energy. How much impulse (in units of N sec) does the mass originally at rest receive during the collision?
Hints: All motion is in 1D. Ignore friction between the masses and the horizontal surface. You will probably need to use the quadratic formula to solve the resulting equations. VR must be greater than VL since the masses can't pass through each other!
Explanation / Answer
let,
mass m1=0.1 kg and m2=0.1 kg
u1=0.29 m/sec and u2=0
use,
m1*u1+m2*u2=m1*v1+m2*v2
0.1*0.29+0=0.1*v1+0.1*v2
0.1*0.29=0.1*v1+0.1*v2
v1+v2=0.29 -----(1)
and
1/2*m*u1^2 +1/2*m2*u2^2 = (1/0.8)*1/2*(m1*v1^2 + m2*v2^2)
1/2*0.1*0.29^2 +0 = (1/0.8)*1/2*0.1*(v1^2+v2^2)
1/2*0.1*0.29^2 = (1/0.8)*1/2*(0.1)*(v1^2+v2^2)
===> v1^2 +v2^2 =0.067 -----(2)
from equation (1) and (2)
v1^2+(0.29-v1)^2=0.067
===> v1=0.26 m/sec
therefore, v2=0.29-v1
v2=0.29-0.26
v2=0.03 m/sec
impulse = change in momentum
=m2*v2 - m2*u2
=0.1*0.03
=3*10^-3 N.sec