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Standard XII

Physics

Law of Conservation of Mechanical Energy

Two blocks of...

Question

Two blocks of M and 3M are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them. A cord initially holding the blocks together is burned; after this, the block of mass 3M moves to the right with a speed of 2.00 m/s.
(a) What is the speed of the block of mass M?
(b) Find the original elastic energy in the spring if M = 0.30 kg.

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Solution

Given,
The initial velocity of both the bodies $V_{i n i t i a l} = 0$
Initial momentum is zero.
Final Velocity of 3M block $V_{B} = 2 m / s$
Mass of block $A = M$
Mass of block $B = 3 M$
Apply conservation of momentum
Final momentum = initial momentum
$3 M V_{B} + M V_{A} = 0$
$V_{A} = -3 V_{B}$ =- $- 3 \times 2 = - 6 m / s$

Given,
Mass $M = 0.30 k g$
Energy in spring = final kinetic Energy of both Masses
$K . E = \frac{1}{2} M V_{A}^{2} + \frac{1}{2} (3 M) V_{B}^{2}$
$K . E = \frac{1}{2} 0.3 \times {(- 6)}^{2} + \frac{1}{2} (3 \times 0.3) 2^{2} = 7.2 J$
(a) Speed of block mass M is $6 m / s$

(b) Total Kinetic Energy is $7.2 J$

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