ckage of mass m is released from rest at a warehouse loading dock and slides dow
ID: 1731917 • Letter: C
Question
ckage of mass m is released from rest at a warehouse loading dock and slides down a Ahuich frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break out having removed the previous package, of mass 2m, from the bottom of the chute as shown in the figure. blacles and use thahe final 3.0 m 2m Conse wetron S) momentum fs'01 a. Suppose the packages stick together. What is their common speed after the collision? ond Hem ve sohe ov man ,then we mz (3m pees z2. b. Suppose the collision between the packages is perfectly elastic. Tohat height does ?ackage of mass m rebound? m?2? 1 21Explanation / Answer
(a) Initial energy of the mass m will be = mgH
where H is the height of the mass m = 3 m , therefore
=m*g*3 -----(1)
Now the energy of this mass when reach at the bottom
= (1/2)mu2 -----------(2)
where u is the velocity at the bottom
Applying energy conservation
(1/2)mu2 = mg*3
u = (2*9.81*3)1/2 = 7.672 m/s
Now the initial momentum of the system before the collision
= mu +(2m)*0 = mu ---------------(3)
Since the collision is inelastic therefore the mass will stick to the truck after the collision , hence
Final momentum after the collision = (2m+m)V = 3mV ----------(4)
where V is the final velocity after the collision
Applying the conservation of momentum
3mV = mU
3V = u
3V = 7.672
V = 2.557 m/s
(b) Now if the collision is elastic then
consider the final velocity of mass m will be V1
final velocity of the truck will be V2
Then the final momentum will be = mV1 + 2mV2 -------------(5)
Applying the conservation of momentum
mV1 + 2mV2 = m u
V1 + 2V2 =7.672 ----------(6)
Now we know that in the elastic collision
Velocity of seperation = velocity of approach
V2 - V1 = u1 - u2
V2 - V1 = u - 0 = 7.672
V2 - V1 = 7.672 ----------(7)
On solving 6 and 7 , we get
V1 = -2.557 m/s
-ve sign shows that it rebound in back direction
V2 = 5.115 m/s
Now the energy of the mass m after the collision = (1/2)mV12
Let say they reach at height h
therefore the final energy = mgh
applying the energy conservation
mgh = (1/2)mV12
h = (1/2g)*V12 = (1/2*9.81)*2.5572
h = 0.333 m
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