A block of mass m 1 = 1 kg slides along a frictionless table with a velocity of

Question

A block of mass m1 = 1 kg slides along a frictionless table with a velocity of +10 m/s. Directly in front of it, and moving with a velocity of +3 m/s, is a block of mass m2 = 5 kg. A massless spring with spring constant k = 1120 N/m is attached to the second block as in the figure below.

1)

(a) Before m1 runs into the spring, what is the velocity of the center of mass of the system?
m/s

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2)

(b) After the collision, the spring is compressed by a maximum amount x. What is the value of x?
cm

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3)

(c) The blocks will eventually separate again. What is the final velocity of each block measured in the reference frame of the table?
m/s (for m1)

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4)

m/s (for m2)

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Explanation / Answer

1) By conservation of momentum, m1u1+m2u2 = (m1+m2)vcm where u1 and u2 are the intial velocities of masses m1 and m2 respectively and vcm is the centre of mass velocity of the system.

1*10+5*3 = (1+5)vcm=> vcm= 25/6 = 4.16m/s

2) The spring is compressed at its maximum amount after the collision when the velocity of the two blocks will be the same and equal to vcm. The kinetic energy that is lost in the collision must be stored in the spring, so,
KE1Before + KE2Before = PEspring + KEafter

1/2(1)102+ 1/2(5) 32= 1/2(1120)x2+ 1/2(6)(4.16)2 => x = 0.1915m

3) Here, we assume that as if the collision was perfectly elastic and hence from momentum conservation,

m1u1+m2u2 = m1v1+m2v2

=> 1.10+5.3 = 1.v1+5.v2 => 25 = v1+5v2----- 1

For elastic collision, relative velocities are related as,

v2-v1 = u1-u2 => v2-v1 = 10-3 = 7--------- 2

Solving 1 and 2, we get, v2 = 5.33m/s and v1 = -1.67m/s

i.e negative sign indicates m1 moves in the opposite direction after collision.

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