Question

m1=6, m2=1.5 collide elastically with each other on a frictionless horizontal track. v1i=1.2, v21=-8. a) v2f=? b)KEtot?

Answer 1

Here we have given that,

m1=6 kg,

m2=1.5kg

v1i=1.2 m/s

v2i= -8 m/s

a) v2f=?

Now using conservation of linear momentum here we will have,

m1V1i + m2V2i = m1v1f + m2v2f

So that here the

V2f = v1i× 2m1/m1+m2 + (m2-m1)/m1+m2 × v2i

On putting the value we will get,

V2f = 1.2× 12/7.5 + (-4.5)×-8 /7.5

V2f = 6.72 m/s

b)now

KEtot will be gievn as ,

Here kinetic energy will be conserved so that inital kinetic energy will be rqual to final kinetic energy.

So that,

KE tot = 0.5 (m1v1i² + m2v2i²) = 52.32 J

Which is our total kinetic energy.