Given,
Let the first mass beΒ m1=m
Let the second mass beΒ m2=4m
Kinetic energy of first mass = Kinetic energy of second mass =Β K.E.
Velocity of first mass =Β V1
Velocity of second mass = V2
Kinetic energy of a particle with massΒ mΒ and velocityΒ vΒ is given by:
K.E=(1x mv^2)/2
Linear momentum of a particle of mass m and velocity V is given by:Β p=mv
We have two cases as the Kinetic energy is equal for both cases.
K.E=(1x mV1^2)/2--------------(i)
K.E=(1x 4m x V2^2)/2--------------(ii)
Dividing equation (i) with (ii) we get,
V1=2V2
β΄Now we have,
p1=m1V1 & p2=m2V2
Dividing P2 by P1 we get
P2/P1= (m2V2)/(m1V1)
P2/P1= (mV2)/(mx 2V2)
P2/P1= 1/2
Hence, the correct answer is option b. 1/2
Given,
Let the first mass beΒ m1=m
Let the second mass beΒ m2=4m
Kinetic energy of first mass = Kinetic energy of second mass =Β K.E.
Velocity of first mass =Β V1
Velocity of second mass = V2
Kinetic energy of a particle with massΒ mΒ and velocityΒ vΒ is given by:
K.E=(1x mv^2)/2
Linear momentum of a particle of mass m and velocity V is given by:Β p=mv
We have two cases as the Kinetic energy is equal for both cases.
K.E=(1x mV1^2)/2--------------(i)
K.E=(1x 4m x V2^2)/2--------------(ii)
Dividing equation (i) with (ii) we get,
V1=2V2
β΄Now we have,
p1=m1V1 & p2=m2V2
Dividing P2 by P1 we get
P2/P1= (m2V2)/(m1V1)
P2/P1= (mV2)/(mx 2V2)
P2/P1= 1/2
Hence, the correct answer is option b. 1/2