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प्रश्न
The blocks shown in figure have equal masses. The surface of A is smooth but that of Bhas a friction coefficient of 0.10 with the floor. Block A is moving at a speed of 10 m/s towards B which is kept at rest. Find the distance travelled by B if (a) the collision is perfectly elastic and (b) the collision is perfectly inelastic.
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उत्तर
Given,
Speed of the block A = 10 m/s
The block B is kept at rest.
Coefficient of friction between the floor and block B, μ = 0.10
Let v1 and v2 be the velocities of A and B after collision, respectively.
(a) If the collision is perfectly elastic, linear momentum is conserved.
Using the law of conservation of linear momentum, we can write:
\[m u_1 + m u_2 = m v_1 + m v_2 \]
\[ \Rightarrow 10 + 0 = v_1 + v_2 \]
\[ v_1 + v_2 = 10 . . . \left( 1 \right)\]
\[\text{ We know,} \]
\[\text{ Velocity of separation } \left( \text{ after collision }\right) = \text{ Velocity of approach } \left( \text{ before collision } \right )\]
\[ v_1 - v_2 = - ( u_1 - v_2 )\]
\[ \Rightarrow v_1 - v_2 = - 10 . . . \left( 2 \right)\]
\[\text{ Substracting equation (2) from (1), we get: }\]
\[2 v_2 = 20\]
\[ \Rightarrow v_2 = 10 \text{ m/s }\]
The deceleration of block B is calculated as follows:
Applying the work-energy principle, we get:
\[\left( \frac{1}{2} \right) \times m \times (0 )^2 - \left( \frac{1}{2} \right) \times m \times v^2 = - m \times a \times s_1 \]
\[ \Rightarrow - \left( \frac{1}{2} \right) \times (10 )^2 = - \mu g \times s_1 \]
\[ \Rightarrow s_1 = \frac{100}{2 \times 0.1 \times 10} = 50 \text{ m }\]
(b) If the collision is perfectly inelastic, we can write:
\[m \times u_1 + m \times u_2 = (m + m) \times v\]
\[ \Rightarrow m \times 10 + m \times 0 = 2m \times v\]
\[ \Rightarrow v = \left( \frac{10}{2} \right) = 5 \text{ m/s }\]
The two blocks move together, sticking to each other.
∴ Applying the work-energy principle again, we get:
\[\left( \frac{1}{2} \right) \times 2m \times (0 )^2 - \left( \frac{1}{2} \right) \times 2m \times (v )^2 = 2m \times \mu g \times s_2 \]
\[ \Rightarrow \frac{(5 )^2}{0 . 1 \times 10 \times 2} = s_2 \]
\[ \Rightarrow s_2 = 12 . 5\]m
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