Question 51: Centre of Mass and Rotational Motion

Question

Consider two blocks A and B of masses $m_1=10 \mathrm{~kg}$ and $\mathrm{m}_2=5 \mathrm{~kg}$ that are placed on a frictionless table. The block A moves with a constant speed $v=3 \mathrm{~m} / \mathrm{s}$ towards the block B kept at rest. A spring with spring constant $\mathrm{k}=3000 \mathrm{~N} / \mathrm{m}$ 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)

Accepted Answer

To solve this problem, we start by using the conservation of linear momentum.
Initially, block A of mass $ m_1 = 10 \, 	ext{kg} $ is moving with velocity $ v_1 = 3 \, 	ext{m/s} $, while block B of mass $ m_2 = 5 \, 	ext{kg} $ is at rest. The blocks move together after the collision. The final velocity of the system is $ v_{	ext{cm}} $. Applying the conservation of linear momentum, we have:
$ m_1 v_1 + m_2 v_2 = (m_1 + m_2) v_{	ext{cm}} $
Substituting the known values:
$ 10 	imes 3 + 5 	imes 0 = (10 + 5) v_{	ext{cm}} $
$ 30 = 15 v_{	ext{cm}} $
Solving for $ v_{	ext{cm}} $:
$ v_{	ext{cm}} = \frac{30}{15} = 2 \, 	ext{m/s} $
Next, we use the conservation of energy to find the compression in the spring. The initial kinetic energy of block A is given by:
$ \frac{1}{2} m_1 v_1^2 = \frac{1}{2} 	imes 10 	imes 3^2 = 45 \, 	ext{J} $
The final kinetic energy of both blocks moving together at $ v_{	ext{cm}} = 2 \, 	ext{m/s} $ is:
$ \frac{1}{2} (m_1 + m_2) v_{	ext{cm}}^2 = \frac{1}{2} 	imes 15 	imes 2^2 = 30 \, 	ext{J} $
The difference in kinetic energy is the energy stored in the compressed spring:
$ \frac{1}{2} k x^2 = 45 - 30 = 15 \, 	ext{J} $
Substituting the given spring constant $ k = 3000 \, 	ext{N/m} $ into the energy equation:
$ \frac{1}{2} 	imes 3000 	imes x^2 = 15 $
$ 1500 x^2 = 15 $
Solving for $ x^2 $:
$ x^2 = \frac{15}{1500} = \frac{1}{100} $
Finally, solving for $ x $:
$ x = \frac{1}{10} = 0.1 \, 	ext{m} $
Therefore, the compression in the spring is $ 0.1 \, 	ext{m} $.

Answer

0 .3 m

Answer

0.1 m

Answer

0.4 m

Answer

0.2 m

Centre of Mass and Rotational Motion

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