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प्रश्न
The friction coefficient between the horizontal surface and each of the block shown in figure is 0.20. The collision between the blocks is perfectly elastic. Find the separation between the two blocks when they come to rest. Take g = 10 m/s2.
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उत्तर
Given:
Initial velocity of 2 kg block, v1 = 1.0 m/s
Initial velocity of the 4 kg block, v2 = 0
Let the velocity of 2 kg block, just before the collision be u1.
Using the work-energy theorem on the block of 2 kg mass:
The separation between two blocks, s = 16 cm = 0.16 m
\[\therefore \left( \frac{1}{2} \right)m \times u_1^2 - \left( \frac{1}{2} \right)m \times (1 )^2 = - \mu \times mg \times s \]
\[ \Rightarrow u_1 = \sqrt{(1 )^2 - 2 \times 0 . 20 \times 10 \times 0 . 16}\]
\[ \Rightarrow u_1 = 0 . 6 \text{ m/s}\]
As the collision is perfectly elastic, linear momentum is conserved.
Let v1, v2 be the velocities of 2 kg and 4 kg blocks, just after collision.
Using the law of conservation of linear momentum, we can write:
\[m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]
\[ \Rightarrow 2 \times 0 . 6 + 4 \times 0 = 2 v_1 + 4 v_2 \]
\[ \Rightarrow 2 v_1 + 4 v_2 = 1 . 2 . . . (1)\]
For elastic collision,
Velocity of separation (after collision) = Velocity of approach (before collision)
\[i . e . v_1 - v_2 = + ( u_1 - u_2 )\]
\[ = + (0 . 6 - 0)\]
\[ \Rightarrow v_1 - v_2 = - 0 . 6 . . . (2)\]
\[\text{ Substracting equation (2) from (1), we get: }\]
\[3 v_2 = 1 . 2\]
\[ \Rightarrow v_2 = 0 . 4 \text{ m/s}\]
\[ \therefore v_1 = - 0 . 6 + 0 . 4 = - 0 . 2 \text{ m/s} \]
Let the 2 kg block covers a distance of S1.
∴ Applying work-energy theorem for this block, when it comes to rest:
\[\left( \frac{1}{2} \right) \times 2 \times (0 )^2 + \left( \frac{1}{2} \right) \times 2 \times (0 . 2 )^2 = - 2 \times 0 . 2 \times 10 \times S_1 \]
\[ \Rightarrow S_1 = 1 cm .\]
Let the 4 kg block covers a distance of S2.
Applying work energy principle for this block:
\[\left( \frac{1}{2} \right) \times 4 \times (0 )^2 - \left( \frac{1}{2} \right) \times 4 \times (0 . 4 )^2 = - 4 \times 0 . 2 \times 10 \times S_2 \]
\[ \Rightarrow 2 \times 0 . 4 \times 0 . 4 = 4 \times 0 . 2 \times 10 \times S_2 \]
\[ \Rightarrow S_2 = 4 \text{ cm}\]
Therefore, the distance between the 2 kg and 4 kg block is given as,
S1 + S2 = 1 + 4 = 5 cm
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