Question

two particles of masses 1g and 4g are moving with equal kinectic energy.what is the ratio of their linear momenta

Answer 1

not true the velocity is not the same, (1/2)*m1*v1^2=(1/2)*m2*v2^2.... this will give you v1 in terms of v2 and you can then use the ratio above

Answer 2

v1=[sqrt(m2/m1)]*v2

Answer 3

oh sorry, I read the question wrong. Yes you are right what I typed above is WRONG!!! I read the velocities were the same. Of course the velcoties are not the same.

Answer 4

The answer is1:2

Answer 5

mA=1g
mB=4g

mB is 4 times larger than mA

EcA=(mAvA^2)/2
EcB=(mBvB^2)/2

EcA=EcB
(mAvA^2)/2=(4mAvB^2)/2
vA^2=4vB^2
vA=2vB

p=linear momentum
pA=mAvA
pB=mBvB

pA/pB=[mAvA]/[mBvB]
pA/pB=[mA2vB]/[4mAvB]
pA/pB=2/4
pA/pB=1/2

Answer 6

k=(1/2)mv^2
k=(1/2m)((mv)^2)
here mv is linear momenta
(mv)^2=2km


(m1v1)^2/(m2v2)^2=2km1/2km2=m1/m2=1/4
therefore, (m1v1)/(m2v2)=1/2
hence,ratio of their linear momenta is 1/2

Answer 7

did u get this?

Answer 8

plz answer my question..........................