Two blocks have mass m and m=2.77 m, respectively. a light spring is attached to one of them, and the blocksare pushed together, compressing the spring between them. they are secured, in this compressed state, by a cord. they are then placed, at rest, on a horizontal frictionless surface. the cord holding them together is burned, after which the block of mass m moves to the right with a speed of 1.38 m/s. what is the speed of the block of mass m? answer in units of m/s

3.8226 m/s

Explanation:

Momentum is conserved hence

equating momentum, p

p (before) = p (after)

0 = (m1v1)+(M2v2)  but m2=2.77m where m and v are the masses and velocity respectively

0 = (mv1) +(2.77m)(1.38)

(mv1)=-(2.77m)(1.38)

(v1)=-(2.77)(1.38)

Solving v1

v1 = - 3.8226 m/s

= 3.8226 m/s, to the left

According to the law of conservation of linear momentum, the linear

momentum of the blocks is conserved (remains the same).

Reasons:

Mass of the two blocks are; m and M = 2.77·m

Location of the spring = Between the blocks

Speed of the block of mass M after the cord is burned, v₂ = 1.38 m/s

Required:

The speed of the block of mass m

Solution:

According to the law of conservation of linear momentum, we have;

Initial momentum of the blocks = Final momentum after the cord is burned

The initial velocity of the blocks when secured, v₀ = 0

v₁ = The velocity of the block of mass m

Therefore;

(m + M) × 0 = m·v₁ + M × 1.38 = m·v₁ + 2.77·m × 1.38 = m·v₁ + 3.8226·m

(m + M) × 0 = 0

Which gives;

(m + M) × 0 = 0 = m·v₁ + 3.8226·m

m·v₁ + 3.8226·m = 0

m·v₁ = -3.8226·m

v₁·m = -3.8226·m

By comparison of the coefficient of m in the above equation, the velocity of the block of mass m, v₁ = -3.8226 m/s.

Therefore, the magnitude of the velocity of block m, |v₁| = 3.8226 m/s

The direction in which the block of mass m moves = opposite to the direction of the block of mass M.

Given that speed is a scalar quantity, we have;

Learn more about the law of conservation of linear momentum here:

link

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