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JEE Advanced 2018 · Paper 1, Physics question 10

A spring-block system is resting on a frictionless floor as shown in the figure. The spring constant is 2.0 N m−12.0\ \mathrm{N\,m^{-1}} and the mass of the block is 2.0 kg. Ignore the mass of the spring. Initially the spring is in an unstretched condition. Another block of mass 1.0 kg moving with a speed of 2.0 m s−12.0\ \mathrm{m\,s^{-1}} collides elastically with the first block. The collision is such that the 2.0 kg block does not hit the wall. The distance, in metres, between the two blocks when the spring returns to its unstretched position for the first time after the collision is ______.

On a frictionless floor, a 1 kg block moves right at 2 m s⁻¹ towards a 2 kg block that is attached by a spring to a wall on the right.
Answer and worked solution

Official answer: 2

Elastic collision of 1 kg at 2 m/s with 2 kg at rest:
v1=1−21+2(2)=−23 m/s,v2=2(1)1+2(2)=43 m/s.v_1=\frac{1-2}{1+2}(2)=-\frac23\text{ m/s},\qquad v_2=\frac{2(1)}{1+2}(2)=\frac43\text{ m/s}.
The 1 kg block moves back at 23\tfrac23 m/s.

Spring. The 2 kg block compresses the spring and returns to the natural-length point after half a period:
T2=πmk=π22=π s.\frac T2=\pi\sqrt{\frac mk}=\pi\sqrt{\frac22}=\pi\text{ s}.

Separation. The 2 kg block is back where the collision happened. In π\pi s the 1 kg block has moved
23π≈2.09 m.\frac23\pi\approx2.09\text{ m}.

Answer: 2.09 (key: 2.00 to 2.20)

Topics:Elastic and inelastic collisionsSpring-mass systems
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