A 15 kg block is firmly attached to a very light horizontal spring (k = 550 N/m) and resting on a smooth horizontal table. Suddenly it is struck horizontally by a 3 kg stone travelling with a velocity of 8 m/s to the right, whereupon the stone rebounds at 2 m/s to the left. Find the maximum distance that the block will compress the spring after the collision.

###### Solution.

Before the collision only the stone has momentum. After the collision both the stone and the block have momentum, but the total momentum should not change. Let’s use the law of momentum conservation and calculate the velocity of the block just after the collision.

m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{2}v_{2f}

Where m_{1} is the mass of the block, m_{2} is the mass of the stone, v_{1i} and v_{1f} are velocities of the block before and after collision respectively, v_{2i} and v_{2f} are velocities of the stone before and after collision respectively.

15 * 0 + 3 * 8 = 15 * v_{1f} – 3 * 2

v_{1f} = 2 m/s

Next we apply the law of conservation of energy

E_{k} = U_{sp}

where E_{k} – the kinetic energy of the block and U_{sp} – the potential energy of the spring.

m_{1} * v_{1f}^{2} / 2 = k * x^{2} / 2

Hence,

15 * 2^{2} = 550 * x^{2}

x ≈ 0.33 m