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marin [14]
3 years ago
6

A railroad car having a mass of 15 Mg is coasting at 1.5 m/s on a horizontal track. At the same time another car having a mass o

f 12 Mg is coasting at 0.75 m/s in the opposite direction. If the cars meet and couple together, determine the speed of both cars just after the coupling. Find the difference between the total kinetic energy before and after coupling has occurred. Explain qualitatively what happened to this energy.
Physics
1 answer:
luda_lava [24]3 years ago
4 0

Answer:

Explanation:

We shall apply law of conservation of momentum to know velocity after collision . Let it be v .

total momentum before collision = total momentum after collision

15 x 1.5 - 12 x .75 = ( 15 + 12 ) v

v = .5 m /s

kinetic energy before collision

1/2 x 15 x 1.5² + 1/2 x 12 x .75²

= 16.875 + 3.375

= 20.25 J

kinetic energy after collision

= 1/2 x ( 15 + 12 ) x .5²

= 3.375 J

Loss of energy = 16.875 J

This energy appear as heat and sound energy that is produced during collision .

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Answer:

C = 17 i^ - 7 j^ + 16 k^ ,   | C| = 24.37

Explanation:

To work the vactor component method, we add the sum in each axis

C = A + B = (Aₓ + Bₓ) i ^ + (A_{y} + B_{y}) i ^ + (A_{z} + B_{z}) k ^

Cₓ = 12+ 5 = 17

C_{y} = -37 +30 = -7

C_{z} = 58 -42 = 16

Resulting vector

C = 17 i ^ - 7j ^ + 16k ^

The mangitude of the vector is

| C | = √ c²

| C | = √( 17² + 7² + 16²)

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3 years ago
Sultan throws a ball horizontally from his window, 12 m above the garden. It reaches the ground after
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Explanation:

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3 years ago
Assuming a 8 kilogram bowling ball moving at 2 m/s bounces off a spring at the same speed that had before bouncing what is the a
Naya [18.7K]

a) 32 kg m/s

Assuming the spring is initially at rest, the total momentum of the system before the collision is given only by the momentum of the bowling ball:

p_i = m u = (8 kg)(2 m/s)=16 kg m/s

The ball bounces off at the same speed had before, but the new velocity has a negative sign (since the direction is opposite to the initial direction). So, the new momentum of the ball is:

p_{fB}=m v_b =(8 kg)(-2 m/s)=-16 kg m/s

The final momentum after the collision is the sum of the momenta of the ball and off the spring:

p_f = p_{fB}+p_{fS}

where p_{fS} is the momentum of the spring. For the conservation of momentum,

p_i = p_f\\p_i = p_{fB}+p_{fS}\\p_{fS}=p_i -p_{fB}=16 kg m/s -(-16 kg m/s)=32 kg m/s


b) -32 kg m/s

The change in momentum of bowling ball is given by the difference between its final momentum and initial momentum:

\Delta p=p_{fb}-p_i=-16 kg m/s - 16 kg m/s=-32 kg m/s


c) 64 N

The change in momentum is equal to the product between the average force and the time of the interaction:

\Delta p=F \Delta t

Since we know \Delta t=0.5 s, we can find the magnitude of the force:

F=\frac{\Delta p}{\Delta t}=\frac{-32 kg m/s}{0.5 s}=-64 N

The negative sign simply means that the direction of the force is opposite to the initial direction of the ball.


d) The force calculated in the previous step (64 N) is larger than the force of 32 N.

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Make I the subject<br> E=VIt
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