Question

Three Velcro blocks, i.e Velcro on their ends, are shown below. The masses and velocities of the blocks are m1 = 7.6 kg, v, = 11.0 m/s. m2 = 10.0 kg, v2 = 4.0 m/s and m3 = 4.0 kg, v3 = 2.0 m/s. Mass m, collides with m2 and the two collide with m3, both collisions perfectly inelastic. The masses then collide with a spring (not shown) with a spring constant of 2.7 times 104 N/m. (a) What is the common velocity of the three masses? (b) How much was the spring compressed when the three masses come to rest against it? (c) What percent of the initial KE was lost?

Answer 1

First Collision

We calculate the final velocity at the first collision:

conservation of linear momentum

For the second collision:

Also apply the conservation of linear momentum:

For the spring we use the conservation of energy:

$\frac{1}{2}(m_1+m_2+m_3)V^2 = \frac{kx^2}{2}$

using given data:

C The lost energy is:

$\Delta E = E_F-E_0 = \frac{kx^2}{2}-\left ( \frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2+\frac{1}{2}m_3v_3^2 \right )$

using given data:

$\Delta E = -238.5 J$