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Two hockey pucks slide along a flat sheet of ice with no friction, and then collide with each other. The diagram below shows a top-down view of the colliding hockey pucks. The masses of the hockey pucks are ma = 2.00 kilograms (kg) and mg = 1.50 kg. Before the collision, puck A moves to the right at a speed of 2.50 meters per second (m/s) and puck moves to the left at a speed of 1.25 m/s. After the collision, puck A moves at a speed of 1.00 m/s at 50.0 degrees relative to the x axis. 1.00 m/s 2.50 m/s 50.0° A 1.25 m/s B before collision after collision A.

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Question

B/c only

Problem 3
Two hockey pucks slide along a flat sheet of ice with no friction, and then collide with
each other. The diagram below shows a top-down view of the colliding hockey pucks.
The masses of the hockey pucks are ma = 2.00 kilograms (kg) and mg = 1.50 kg. Before
the collision, puck A moves to the right at a speed of 2.50 meters per second (m/s) and
puck moves to the left at a speed of 1.25 m/s. After the collision, puck A moves at a speed
of 1.00 m/s at 50.0 degrees relative to the x axis.
1.00
m/s
2.50
m/s
50.0°
1.25
m/s
A
before collision
after collision
(a) Find find the final speed (in m/s) of puck B after the collision.

(b) Find find the direction of travel of puck B after the collision. Give your
answer as an angle in degrees, measured relative to the x axis.
(c) What is the total change in kinetic energy of the hockey pucks during the
collision, in units of Joules (J)?

Expert Solution

Step 1

This is an example of inelastic collision.

In in-elastic collisions, the momentum of the colliding bodies is conserved if no other forces act on them, but their kinetic energies are not conserved.

Step 2

Both the pucks have their initial velocities along the x direction. And since momentum is conserved, the momentum conservation in the x direction is

$p_{1 x} - p_{2 x} = {p_{1 x}}^{'} \cos (θ_{1}) + {p_{1 x}}^{'} \cos (θ_{2}) S i n c e, t h e i n i t i a l m o m e n t a f o r b o t h p u c k s a r e o p p o s i t e l y d i r e c t e d A n d a l o n g t h e x - a x e s, t h e i r m o m e n t u m c o m p o n e n t s a r e a l o n g t h e s a m e d i r e c t i o n s m_{1} v_{1 x} - m_{2} v_{2 x} = m_{1} {v_{1 x}}^{'} \cos (50) + m_{2} {v_{2 x}}^{'} \cos (θ_{2}) (2 \times 2.5) - (1.25 \times 1.5) = (2 \times 1) 0.6428 + 1.5 {v_{2 x}}^{'} \cos (θ_{2}) 3.125 = 1.2856 + 1.5 {v_{2 x}}^{'} \cos (θ_{2}) 3.125 - 1.2856 = 1.5 {v_{2 x}}^{'} \cos (θ_{2}) 1.5 {v_{2 x}}^{'} \cos (θ_{2}) = 1.8394$

Now, along the y-direction

$T h e i n i t i a l m o m e n t u m f o r b o t h t h e p u c k s a l o n g t h e y - a x e s i s 0 . S o, 0 = m_{1} {v_{1 y}}^{'} \sin (θ_{1}) - m_{2} {v_{2 y}}^{'} \sin (θ_{2}) T h e v e r t i c a l c o m p o n e n t s o f t h e v e l o c i t i e s f o r b o t h t h e p u c k s a r e o p p o s i t e l y d i r e c t e d m_{1} {v_{1 y}}^{'} \sin (θ_{1}) = m_{2} {v_{2 y}}^{'} \sin (θ_{2}) (2 \times 1) \sin (50) = 1.5 {v_{2 y}}^{'} \sin (θ_{2}) 2 \times 0.766 = 1.5 {v_{2 y}}^{'} \sin (θ_{2}) 1.5321 = 1.5 {v_{2 y}}^{'} \sin (θ_{2})$

Step 3

$N o w, \frac{1.5 {v_{2 y}}^{'} \sin (θ_{2})}{1.5 {v_{2 x}}^{'} \cos (θ_{2})} = \frac{1.5321}{1.8394} \tan (θ_{2}) = 0.83293 θ_{2} = \tan^{- 1} (0.83293) θ_{2} = 39.792 °$

This is the direction that the puck B makes with the x-axis after collision.

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