simple pendulum consisting of a 5.0kg mass and a rod oflength 1.6m (rod has no m
ID: 2068626 • Letter: S
Question
simple pendulum consisting of a 5.0kg mass and a rod oflength 1.6m (rod has no mass). The pendulum is released fromrest at a 20 deg angle below the horizontal. When thependulum is vertically straight down, it collides with a 2.0kg massthat is traveling straight horizontally to the right with avelocity of 2.00 m/s. After the collision, the 2.0kg masstravels .425m to the left before coming to a stop. If thecoefficient of kinetic friction between the 2.0kg mass andhorizontal surface is .300, find the following:a) the maximum vertical height of the 5.0kg mass and thecorresponding angle the rod makes with the vertical after thecollision between the masses.
b) the work done by gravity on the5.0kg mass as it travels from the collision point to its maximumheight.
Explanation / Answer
Initial potential energy of 5 kg mass is 5*9.81*(1.6 - 1.6*Sin20) = 51.64 J
K.E of 5 kg mass at time of impact = 1/2*5*v2
Equating both, 51.64 = 1/2*5*v2 So, v = 4.545 m/s
This will be the velocity of 5 kg mass at the time of impact.
Doing linear momentum conservation before and after the collision,
5*4.545 + 2*(-2) = 2*v2 + 5*v5 where v2 and v5 represent the velocities of 2 kg and 5 kg masses after the collision respectively.
Friciton force on 2 kg mass = 0.3*(2*9.81) = 5.886 N
Deceleration of 2 kg mass = Friction force/m = 5.886/2 = 2.943 m/s2
Using v2 = u2 - 2as we get, 0 = v22 - 2*2.943*0.425
Thus v2 = 1.58 m/s
So, v5 = 3.11 m/s
So, KE of 5 kg mass after collision = 1/2*5*3.112 = 24.22 J
(a) Potential energy of 5 kg mass when it reaches its max. height h is 5*9.81*h.
Using energy conservation, 5*9.81*h = 24.22
So, h = 0.494 m
Taking angle between rod and vertical to be , 1.6 - 1.6*Cos = 0.494
This gives = 46.3 degrees.
(b) Work done by gravity = potential energy of 5 kg mass = 24.22 J
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