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`        A particle of mass 4.0 kg, initially moving with a velocity of 2.0 m/s collides elastically with a particle of mass 6.0 kg. initially moving with a velocity of -4.0 m/s. What are the velocities of the two particles after the collision?`
9 months ago

Raghav Singh
20 Points
```							Take right hand side to be positive(+) and left hand side to be negative(-)  Initially, Speed of 4kg block= 2m/s (left to right) and Speed of 6kg block= 4m/s (right to left)Let speed after collision to be v1 and v2 respectively (left to right both) Conserving momentum we have Pinitial = Pfinal4(2) – 6(4) = 4(v1) + 6(v2) -16 = 4(v1) + 6(v2)2(v1) + 3(v2) = -8 ----equation (i) Now as collison is elastic e=1, i.e1= velocity of separation after collision/velocity of approach before collision1= (v2 – v1)/(4+2)6= v2 – v1 i.e2(v2) – 2(v1) = 12 ------equation (ii) adding equations (i) and (ii) we have5(v2) = 4v2 = 4/5 (as it is +ve it means it is moving from left to right)Putting value of v2 in equation (ii) we get v1 = -26/5 (as it is -ve it means it is moving from right to left)
```
9 months ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions