A mass of 0.20 kg is suspended from the lower end of a light spring. A second mass of 0.10 kg is suspended from the first mass by a thread.

Question 5

A mass of 0.20 kg is suspended from the lower end of a light spring. A second mass of 0.10 kg is suspended from the first mass by a thread. The arrangement is allowed to come into static equilibrium and then the thread is burned through.

At this instant, what is the upward acceleration of the 0.20 kg mass? (Assume g = 10 m s^-2.)

A 5.0 m s^-2        B 6.7 m s^-2                        C 10 m s^-2                         D 15 m s^-2

Reference: Past Exam Paper – November 2015 Paper 12 Q12

Solution:

Answer: A.

Before the thread is burned,

Downward force on the larger mass = 2.0 N weight (its own weight) + 1.0 N weight of

                                                       smaller mass).

Downward force on the larger mass = 3.0 N

From Newton’s 3^rd law, the spring exerts an equal upward force.

When the thread is burned, the smaller mass is released.

The downward force on the larger mass is now due to its own weight only (= 2.0 N). The upward force is still 3.0 N at this instant.

There is an instantaneous resultant force of (3.0 – 2.0 =) 1.0 N upwards on the 0.20 kg mass.

Resultant force = ma = 1.0 N

Acceleration a = 1.0 / 0.2 = 5.0 m s^-2