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Answer:
3.7 m/s
Explanation:
M = 444 kg
U = 5 m/s
m = 344 kg
u = - 5 m/s
Let the velocity of train is V and the car s v after the collision.
As the collision is elastic
By use of conservation of momentum
MU + mu = MV + mv
444 x 5 - 344 x 5 = 444 V + 344 v
500 = 444 V + 344 v
125 = 111 V + 86 v .... (1)
By using the formula of coefficient of restitution ( e = 1 for elastic collision)

-5 - 5 = V - v
V - v = - 10
v = V + 10
Substitute the value of v in equation (1)
125 = 111 V + 86 (V + 10)
125 = 197 V + 860
197 V = - 735
V = - 3.7 m/s
Thus, the speed of first car after collision is 3.7 m/s. negative sign shows that the direction is reverse as before the collision.
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<span>3) Neither precise or accurate.
This is because of the deviation between the measurements, they vary and are not within a good range. And they are not close to the accepted value. In order to be precise the measurements have to be relatively close to each other, and to be accurate they have to be close to the accepted value.</span>
Answer:
The magnitude of the velocity is approximately 11.45 units
Explanation:
Since they give us the x and y components of the vector velocity, we can find its magnitude via the Pythagorean theorem:
