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Question

A particle of mass m, kinetic energy K and linear momentum p collides head on elastically with another particle of mass 2m at rest. Match the following (after collision) –

Column-IColumn-II(A) momentum of first particle(P) 43p(B) momentum of second particle(Q) 8K9(C) Kinetic energy of first particle(R)p3(D) Kinetic energy of second particle(S)K9

Which of the following option has the correct combination considering column-I and column-II

A

CS
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B

DP,R
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C

BQ
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D

AP,Q
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Solution

The correct option is A
CS
Let m1 and m2 be the mass of two collision bodies.

u1 and u2 be their velocities before collisions v1 and v1 be their velocities after collisions
In the given situation,
m1=m ;m2=2mu1=u1 ; u2=0
Using conservation of momentum and energy we have equations.

v1=(m1m2m1+m2)u1+(2m1m1+m2)u2 ...(1)

v2=(m2m1m1+m2)u2+(2m1m1+m2)u1 ...(2)
Putting values given in question in equation (1) and (2) ; We get:
v1=u13 & v2=23u1
According to question
P=m1v1 & K=12m1v21
Final momentum of mass m1
m1v1=mu13=P3

Final K.E of mass m1;
12m1v21=12m1×u219=K9

Final Momentum of mass m2
m2v2=2mv2=2m×23u1=43P
Final KE of mass m212m2v2212×2m×(49u21)8K9

Matching the option we
(AR) (BP) (CS) (DQ)

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