(1) Puck 1 has a mass of 146g; puck 2 has a mass of 48g. The centers of the puck
ID: 1768048 • Letter: #
Question
(1) Puck 1 has a mass of 146g; puck 2 has a mass of 48g. The centers of the pucks are 20 cm apart. At what position is the center of mass located, relative to the center of puck 1?
(2) Assume that for the conditions before the collision, you measured the following
the x-component of the velocity of puck 1 = +0.147 m/s
the x-component of the velocity of puck 2 = +0.370 m/s
the y-component of the velocity of puck 1 = +0..297 m/s
the y-component of the velocity of puck 2 = -0.368 m/s
Using the same masses as in Question 1, calculate the following:
(a) x-component of the momentum of puck 1
(b) x-component of the momentum of puck 2
(c) x-component of the momentum of the system
(d) y-component of the momentum of puck 1
(e) y-component of the momentum of puck 2
(f) y-component of the momentum of the system
(g) magnitude of the momentum of the system
Explanation / Answer
Center of mass from puck 1=m2*R/(m1+m2)
=48*20/(146+48)
=4.948 cm from pluk 1
x-component of the Momentum of pluk 1=m1*v1x=146*10^-3*0.147 =21.46 *10^-3 kg.m/s
x-component of the Momentum of pluk 2= m2*v2x = 0.048*0.37 =17.76 *10^-3 kg.m/s
x-component of the momentum of the system= p1x + p2x =49.22 *10^-3 kg.m/s
y-component of the Momentum of pluk 1= m1*v1y = 0.146*0.297 = 43.36 *10^-3 kg.m/s
y-component of the Momentum of pluk 2= m2*v2y = -0.048*0.368 = -17.66 *10^-3 kg.m/s
y-component of the momentum of the system=25.69 *10^-3 kg.m/s
magnitude of the momentum of the system=55.5 *10^-3 kg.m/s
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