Question

A block of mass 2kg is released from a height 10m on to a smooth sledge of mass 8kg filled with an ideal spring of force constant K = 100 N/m. Choose the correct options. (g = 10 m/s²)

• A. Velocity of the block just before block touches the spring is $4\sqrt{10}$ m/s.
• B. Velocity of sledge just before block touches the spring is $\sqrt{10}$ m/s.
• C. Maximum work done by spring on the system consisting block and sledge during subsequent motion is 200 J.
• D. Maximum compression in the spring is 20cm.

Answer 1

Answer: (A), (B), (C)

A, B, C

Answer 2

The problem describes a block sliding down a smooth ramp onto a smooth sledge containing a spring. We need to analyze the motion and evaluate the given options.

Given values:

• Mass of block, $m_1 = 2 \text{ kg}$
• Mass of sledge, $m_2 = 8 \text{ kg}$
• Height from which block is released, $h = 10 \text{ m}$
• Spring constant, $K = 100 \text{ N/m}$
• Acceleration due to gravity, $g = 10 \text{ m/s}^2$

Analysis of the motion:

Phase 1: Block slides down the ramp on the sledge. Consider the system consisting of the block and the sledge. Since all surfaces are smooth, there are no external horizontal forces acting on this system. Therefore, the horizontal momentum of the system is conserved. Initially, both the block and the sledge are at rest (assuming the sledge is initially at rest and the block is released from a height on it). So, the initial horizontal momentum is zero. As the block slides down, it gains horizontal velocity, and due to the interaction with the ramp (part of the sledge), the sledge also gains horizontal velocity in the opposite direction to conserve momentum.

Let $v_1$ be the speed of the block and $v_2$ be the speed of the sledge just before the block touches the spring. Let's assume the block moves to the right (positive direction) and the sledge moves to the left (negative direction). Conservation of horizontal momentum: $m_1(0) + m_2(0) = m_1v_1 + m_2(-v_2)$ $0 = 2v_1 - 8v_2$ $2v_1 = 8v_2 \Rightarrow v_1 = 4v_2$ (Equation 1)

Also, mechanical energy is conserved for the system (block + sledge + Earth) because there is no friction and the spring is not yet compressed. Initial potential energy of the block: $PE_i = m_1gh$ Initial kinetic energy of the block and sledge: $KE_i = 0$ Final potential energy (at the horizontal level): $PE_f = 0$ Final kinetic energy of the block and sledge: $KE_f = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$

Conservation of energy: $PE_i = KE_f$ $m_1gh = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$ Substitute the given values: $2 \times 10 \times 10 = \frac{1}{2}(2)v_1^2 + \frac{1}{2}(8)v_2^2$ $200 = v_1^2 + 4v_2^2$ (Equation 2)

Now, substitute $v_1 = 4v_2$ from Equation 1 into Equation 2: $200 = (4v_2)^2 + 4v_2^2$ $200 = 16v_2^2 + 4v_2^2$ $200 = 20v_2^2$ $v_2^2 = \frac{200}{20} = 10$ $v_2 = \sqrt{10} \text{ m/s}$

Now find $v_1$: $v_1 = 4v_2 = 4\sqrt{10} \text{ m/s}$

Let's evaluate options (A) and (B): (A) Velocity of the block just before block touches the spring is $4\sqrt{10}$ m/s. This matches our calculated $v_1$. So, option (A) is correct. (B) Velocity of sledge just before block touches the spring is $\sqrt{10}$ m/s. This matches our calculated $v_2$. So, option (B) is correct.

Phase 2: Block compresses the spring. After the block touches the spring, the spring starts to compress. Maximum compression ($x_{max}$) occurs when the relative velocity between the block and the sledge becomes zero, meaning they move with a common velocity, $V_c$. During the compression, the horizontal momentum of the system (block + sledge) is still conserved, as there are no external horizontal forces. Initial horizontal momentum (just before compression): $P_i = m_1v_1 - m_2v_2 = 2(4\sqrt{10}) - 8(\sqrt{10}) = 8\sqrt{10} - 8\sqrt{10} = 0$. Final horizontal momentum (at maximum compression): $P_f = (m_1+m_2)V_c$ Since $P_i = P_f$, we have $0 = (2+8)V_c \Rightarrow 10V_c = 0 \Rightarrow V_c = 0$. This means at maximum compression, the entire system (block + sledge) momentarily comes to rest.

Now, apply the conservation of mechanical energy for the system (block + sledge + spring). Initial energy (just before spring compression begins): $E_{initial} = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$ $E_{initial} = \frac{1}{2}(2)(4\sqrt{10})^2 + \frac{1}{2}(8)(\sqrt{10})^2$ $E_{initial} = (16 \times 10) + (4 \times 10) = 160 + 40 = 200 \text{ J}$ This $200 \text{ J}$ is the total kinetic energy of the system just before the spring starts compressing.

Final energy (at maximum compression $x_{max}$): $E_{final} = \frac{1}{2}(m_1+m_2)V_c^2 + \frac{1}{2}Kx_{max}^2$ Since $V_c = 0$, $E_{final} = 0 + \frac{1}{2}Kx_{max}^2 = \frac{1}{2}Kx_{max}^2$.

By conservation of energy: $E_{initial} = E_{final}$ $200 = \frac{1}{2}Kx_{max}^2$ $200 = \frac{1}{2}(100)x_{max}^2$ $200 = 50x_{max}^2$ $x_{max}^2 = \frac{200}{50} = 4$ $x_{max} = \sqrt{4} = 2 \text{ m}$

Let's evaluate option (D): (D) Maximum compression in the spring is 20cm. Our calculated $x_{max} = 2 \text{ m}$, which is $200 \text{ cm}$. So, option (D) is incorrect.

Let's evaluate option (C): (C) Maximum work done by spring on the system consisting block and sledge during subsequent motion is 200 J. The maximum energy stored in the spring is $PE_{spring, max} = \frac{1}{2}Kx_{max}^2 = 200 \text{ J}$. When the spring is compressed, it does negative work on the system. The magnitude of this work is $200 \text{ J}$. When the spring expands from its maximum compression (2 m) back to its natural length (0 m compression), it does positive work on the system. The work done by the spring during expansion is $W_{spring, expansion} = \int_{x_{max}}^0 F_{spring} dx = \int_{x_{max}}^0 (-Kx) dx = -\frac{1}{2}K[x^2]_{x_{max}}^0 = -\frac{1}{2}K(0 - x_{max}^2) = \frac{1}{2}Kx_{max}^2 = 200 \text{ J}$. This is the maximum positive work the spring can do on the system. Therefore, option (C) is correct.

Final Check: Options (A), (B), and (C) are correct.