Question

Consider two blocks A and B of masses \( m_1 = 10 \) kg and \( m_2 = 5 \) kg that are placed on a frictionless table. The block A moves with a constant speed \( v = 3 \) m/s towards the block B kept at rest. A spring with spring constant \( k = 3000 \) N/m is attached with the block B as shown in the figure. After the collision, suppose that the blocks A and B, along with the spring in constant compression state, move together, then the compression in the spring is, (Neglect the mass of the spring)

• A. 0.2 m
• B. 0.4 m
• C. 0.1 m
• D. 0.3 m

Hint:

In collisions where objects stick together, use the conservation of linear momentum to find the common final velocity. The kinetic energy lost during such inelastic collisions is often converted into other forms of energy, such as potential energy stored in a spring. Use the conservation of energy to relate the loss in kinetic energy to the potential energy stored in the spring and solve for the compression.

Answer 1

The Correct Option is C

We are given two blocks A and B with masses \( m_1 = 10 \) kg and \( m_2 = 5 \) kg on a frictionless table. Block A moves with a velocity \( v_1 = 3 \) m/s towards block B, which is initially at rest (\( v_2 = 0 \) m/s). After the collision, the blocks A and B move together with a common velocity \( v_{cm} \). We can find this common velocity using the principle of conservation of linear momentum: \[ m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_{cm} \] \[ (10 \, \text{kg})(3 \, \text{m/s}) + (5 \, \text{kg})(0 \, \text{m/s}) = (10 \, \text{kg} + 5 \, \text{kg}) v_{cm} \] \[ 30 \, \text{kg m/s} = (15 \, \text{kg}) v_{cm} \] \[ v_{cm} = \frac{30}{15} = 2 \, \text{m/s} \] The kinetic energy lost during the inelastic collision is stored as potential energy in the compressed spring. We can use the principle of conservation of energy. The initial kinetic energy of the system is the kinetic energy of block A: \[ KE_i = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} (10 \, \text{kg}) (3 \, \text{m/s})^2 = \frac{1}{2} (10)(9) = 45 \, \text{J} \] The final kinetic energy of the combined mass (\( m_1 + m_2 \)) moving with velocity \( v_{cm} \) is: \[ KE_f = \frac{1}{2} (m_1 + m_2) v_{cm}^2 = \frac{1}{2} (15 \, \text{kg}) (2 \, \text{m/s})^2 = \frac{1}{2} (15)(4) = 30 \, \text{J} \] The loss in kinetic energy is stored as the potential energy in the compressed spring: \[ PE_{spring} = KE_i - KE_f = 45 \, \text{J} - 30 \, \text{J} = 15 \, \text{J} \] The potential energy stored in a compressed spring with spring constant \( k \) and compression \( x \) is given by: \[ PE_{spring} = \frac{1}{2} k x^2 \] We are given \( k = 3000 \) N/m. So, \[ 15 = \frac{1}{2} (3000) x^2 \] \[ 15 = 1500 x^2 \] \[ x^2 = \frac{15}{1500} = \frac{1}{100} \] \[ x = \sqrt{\frac{1}{100}} = \frac{1}{10} \, \text{m} = 0.1 \, \text{m} \] The compression in the spring is 0.1 m.