
In elastic collision energy of system remains same. so, (KE)before collision =(KE)after collision Let speed of second body after collision is v′. 21Mv2+0=21M(3v)2+21M(v′)2⇒v′=322v
NEET UG 2015 — Physics Mechanics
On a frictionless surfaces, a block of mass M moving at speed v collides elastically with another block of same mass M which is initially at rest. After collision the first block moves at an angle θ to its initial direction and has a speed 3v . The second block's speed after the collision is:
Held on 30 Apr 2015 · Verified 9 Jul 2026.
43v
23v
23v
322v
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