Consider the collision shown in Fig. 2. A puck of mass m1 = 0.50 kg moving to th

Question

Consider the collision shown in Fig. 2. A puck of mass m1 = 0.50 kg moving to the right collides with a stationary puck of mass m2 = 0.30kg. In the graph above, only the track of mi is shown. Consider each point to be a picture of the puck with 30 pictures taken each second. What is the initial momentum of m1? Express it in the unit vector notation. What is the final momentum of m1? Express it in the unit vector notation. What is the final momentum of m2? Express it in both the unit vector and the magnitude and orientation notation.

Explanation / Answer

Frame rate = 30 frames/sec

time between two frames = 1/30 = 0.03333 sec

If we see the mass m1 before collission,

Lets take point upto 3cm

Time taken to cover 3cm = 0.03333*3

= 0.1 sec

Velocity = distance/tme = (3*10^-2) / (0.1)

= 0.3 m/sec

As the velocity is aligned only in x-direction, so in vector notation

V = 0.3i

After collision,

Lets see m1 from 4cm to 7cm,

6 time spans are there between frames

So, velocity in x-direction = (7 - 4)*10^-2 / (6*0.03333)

= 0.15 m/sec

Now, concentrating on y component of velocity

In 6 time frames it covered 1.5 cm in negative y direction

So, velocity in y-direction = (1.5*10^-2) / (6*0.03333)

= 0.075 m/sec

In vector notation

V = 0.15i - 0.075j

As momentum is conserved,

Conserving momentum in y-direction, which initially was zero (it will be zero after collision as well)

0.5*0.075 = 0.3*v

v = 0.125 m/sec in positive y direction

Conserving momentum in x direction

0.5*0.3 = 0.5*0.15 + 0.3*v (m1*v1) = (m1*v1x) + (m2*v2x)

v = 0.25 m/sec in positive x-direction

So, velocity of m2 in vector notation is

V = 0.25i +0.125j

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