Question

two smooth tracks of equal length have bumps a up and b down both of the same curvature and size ....

Answer 1

The speed of the ball at the top of the curve on track A is < 2 m/s

Therefore, option (C) is correct.

Explanation:

Given:

The initial speed of the balls = 3 m/s

The speed of the ball at the bottom of the curve on track B = 4 m/s

To find out:

The speed of the ball at the top of the curve on track A

Solution:

The curvature and size of the curves is same

Therefore, on track A the height ascended at the top of the curve = height descended at the bottom of the curve on track B

Thus,

Gain in kinetic energy on reaching the bottom of the curve on track B = Loss in kinetic energy on reaching the top of the curve on track A

Assuming the masses of both the balls to be same and equal to m

We know that for a particle of mass m and speed v, the kinetic energy is given by

$KE=\frac{1}{2}mv^2$

Thus, if the speed of the ball at the top of the curve on track A is v then

$\frac{1}{2}m\times 4^2-\frac{1}{2}m\times 3^2=\frac{1}{2}m\times 3^2-\frac{1}{2}m\times v^2$

$\implies 16-9=9-v^2$

$\implies v^2=2$

$\implies v=\sqrt{2}$ m/s

Therefore, option (c) is correct.

Hope this answer is helpful.

Know More:

Q: A smooth body is released from rest A at a point A at the top of a smooth curved track of vertical height 40 cm, what is the speed of the body at d bottom of the curved track. How far along the adjoining smooth inclined plane will d body go?

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