Speed of 4kg mass along line of impact after collision is x 1 .Write value of x

Along the line of impact 2 kg sphere is having velocity 2 3 m / s and 4 kg sphere is having velocity − 3 m / s . Along the line impact 2 × 2 3 − 4 3 = 2 V 1 + 4 V 2 V 1 + 2 V 2 = 0 ( 1 ) e = V 1 − V 2 V 2 − V 1 = 2 3 + 3 V 2 − V 1 ⇒ V 2 − V 1 = 3 ( 2 ) From (1) and (2) V 2 = 3 1 m / s ; V 1 = − 3 2 m / s Maximum potential energy of deformation is loss in kinetic energy during deformation along the line of impact U = 2 1 ( 2 ) ( 2 3 ) 2 + 2 1 ( 4 ) ( 3 ) 2 − 2 1 ( 6 ) 2 V CM = 12 + 6 = 18 J .

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Passage / Comprehension

Two smooth spheres are moving towards each other. Both have same radius but their masses are 2 kg and 4 kg. If the velocities are 4 m/s and 2 m/s, respectively and coefficient of restitution is e = 1/3 . Their lines of motion are parallel to each other but have separation equal to the radius of sphere.

Speed of 4kg mass along line of impact after collision is $\frac{1}{x}$ .Write value of $x$

Answer:

1.73

Solution:

Along the line of impact 2 kg sphere is having velocity $23 m / s$ and 4 kg sphere is having velocity $- 3 m / s$ . Along the line impact $2 \times 23 - 43 = 2 V_{1} + 4 V_{2}$

$V_{1} + 2 V_{2} = 0 (1)$

$e = \frac{V _{2} - V _{1}}{V _{1} - V _{2}} = \frac{V _{2} - V _{1}}{2 3 + 3} \Rightarrow V_{2} - V_{1} = 3 (2)$

From (1) and (2) $V_{2} = \frac{1}{3} m / s; V_{1} = - \frac{2}{3} m / s$

Maximum potential energy of deformation is loss in kinetic energy during deformation along the line of impact $U = \frac{1}{2} (2) (23)^{2} + \frac{1}{2} (4) (3)^{2} - \frac{1}{2} (6)^{2}_{V CM} = 12 + 6 = 18 J$ .

Stream:JEESubject:PhysicsTopic:Center of MassSubtopic:Collision

⏱ 2mℹ️ Source: QB

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