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)
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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.
Two blocks, A and B, with masses \( m_1 = 10 \) kg and \( m_2 = 5 \) kg, respectively, are on a frictionless table. Block A moves at \( v_1 = 3 \) m/s towards block B, which is stationary (\( v_2 = 0 \) m/s). Following a collision, both blocks move together with a common velocity \( v_{cm} \). This common velocity is determined using the conservation of linear momentum principle:\[m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_{cm}\]Substituting the given values:\[(10 \, \text{kg})(3 \, \text{m/s}) + (5 \, \text{kg})(0 \, \text{m/s}) = (10 \, \text{kg} + 5 \, \text{kg}) v_{cm}\]\[30 \, \text{kg m/s} = (15 \, \text{kg}) v_{cm}\]Solving for \( v_{cm} \):\[v_{cm} = \frac{30}{15} = 2 \, \text{m/s}\]During the inelastic collision, the lost kinetic energy is converted into potential energy within the compressed spring. The principle of conservation of energy is applied. The initial kinetic energy of the system is that 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 at \( v_{cm} \) 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 kinetic energy lost is stored as 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 is given by \( PE_{spring} = \frac{1}{2} k x^2 \), where \( k \) is the spring constant and \( x \) is the compression. Given \( k = 3000 \) N/m:\[15 = \frac{1}{2} (3000) x^2\]\[15 = 1500 x^2\]\[x^2 = \frac{15}{1500} = \frac{1}{100}\]\[x = \sqrt{\frac{1}{100}} = \frac{1}{10} \, \text{m} = 0.1 \, \text{m}\]The spring's compression is 0.1 m.
