Question.
Mass 1 hits mass 2 and causes the two masses to
fall on the floor as shown in the figure. The surfaces
are NOT frictionless.
a.
Using conservation of energy with consideration of the dissipated energy due to friction, calculate the velocity and momentum just before collision.
b.
Calculate the momentum of mass1 and mass2 after collision.
Answer.
Let u and v be the velocity of mass 1 before and after collision respectively.
Let v' be the velocity of mass 2 after collison. Initially, mass 2 is at rest.
First we calculate v and v' from the projectile motion of the two masses.
Time taken by both the masses to reach the ground
= √(0.7/4.9) s.
= 0.378 s.
In this time mass 1 covers a distance of 0.2 m with velocity v and mass 2 covers a distance of 1.1 m with velocity v'
So, v = 0.2 / 0.378 = 0.53 m/s
and v' = 1.1 / 0.378 = 2.91 m/s
Now, by momentum balance,
25u + 0 = (25)*(0.53) + (15)*(2.91)
=> u = (13.25 + 43.65) / 25 = 2.276 m/s
Dissipated energy due to friction
= P.E. at top - K.E. at the bottom of curved surface
= (0.025)*(9.8)*(0.3) - (1/2)*(0.025)*(2.276^2) joule
= (0.025)*[ (9.8)*(0.3) - (0.5)*(2.276^2) ]
= (0.025)*[ 2.94 - 2.59 ]
= 0.00875 joule
a)
Velocity of mass 1 before collision = 2.276 m/s
Momentum of mass 1 before collision
= (0.025)*(2.276) = 0.0569 (kg)(m)/s
Velocity and momentum of mass 2 before collision = zero.
b)
Momentum of mass 1 after collision
= (0.025)*(0.53) =0.0133 (kg)(m)/s
Momentum of mass 2 after collision
= (0.015)*(2.91) = 0.04365 (kg)(m)/s
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