SATHEE: Rotational Motion Question 74

Rotational Motion Question 74

Question: A solid cylinder of mass 3 kg is rolling on a horizontal surface with velocity $4 m s^{- 1}$ . It collides with a horizontal spring of force constant $200 N m^{- 1}$ . The maximum compression produced in the spring will be

[AIPMT (S) 2012]

Options:

A) 0.5 m

B) 0.6 m

C) 0.7 m

D) 0.2 m

Show Answer

Answer:

Correct Answer: B

Solution:

Loss in $K E =$ Gain in spring energy

$\frac{1}{2} m v^{2} [1 + \frac{K^{2}}{R^{2}}] = \frac{1}{2} k x_{max}^{2}$

where k is the force constant.

Given, $v = 4 m / s, m = 3 k g, k = 200 N / m$

For solid cylinder, $\frac{K^{2}}{R^{2}} = \frac{1}{2}$

$∴$ $\frac{1}{2} \times 3 \times {(4)}^{2} [1 + \frac{1}{2}] = \frac{1}{2} \times 200 \times x_{max}^{2}$

The maximum compression in the spring

$x_{max}^{} = 0.6 m$

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Rotational Motion Question 74

Question: A solid cylinder of mass 3 kg is rolling on a horizontal surface with velocity 4ms−1 . It collides with a horizontal spring of force constant 200Nm−1 . The maximum compression produced in the spring will be

Options:

Answer:

Solution:

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Question: A solid cylinder of mass 3 kg is rolling on a horizontal surface with velocity $4 m s^{- 1}$ . It collides with a horizontal spring of force constant $200 N m^{- 1}$ . The maximum compression produced in the spring will be