A cart with mass 210 g moving on a frictionless linear air track at an initial s

Question

A cart with mass 210 g moving on a frictionless linear air track at an initial speed of 1.4 m/s strikes a second cart of unknown mass at rest. The collision between the carts is elastic. After the collision, the first cart continues in its original direction at 0.91 m/s. (a) What is the mass of the second cart ( g )?
(b) What is its (second cart) speed after impact?
(c) What is the speed of the two-cart center of mass?

This is due later tonight and it does not seem difficult, but I am having a lot of trouble. I will answer any help lifesaver. Thank you!

Explanation / Answer

first cart mass m1= 210gm=0.21kg ,

intial speed of the first cart = u1= 1.4 M/S

intial velocity of the second cart= u2= 0 m/s (it is in rest intially)

after the collision takes place , final velocity of the first cart v1=0.91 m/s

mass of the second cart m2 = ?

velocity of the second cart after the collision  takes place v2=?

acc. to conservation of linear momentum ,

m1u1+m2u2 =m1v1+m2v2

m1(u1-v1)=m2(v2-u2) ------(1)

acc. to conservation of energy ,

1/2m1u1^2 + 1/m2u2^2 = 1/2mv1^2+ 1/2m2v2^2

m1(u1^2- u2^2) =m2(v2^2-v1^2)------------(2)

on dividing 1/2

u1+v1=v2+u2       ---------(3)          (only for elstic collision)

given u1= 1.4 m/s, v1=0.91m/s , u2=0 m/s , v2=?

      v2= u1+v1-(u2)

         v2 =1.4+0.91=2.31 m/s

ans for (b) is 2.31 m/s

sub. the value of v2 from the eqn. (3)   in eqn (1) then ,

we will have

v1= (m1-m2/m1+m2)u1 +2m2u2/m1+m2

but u2=0 m/s then,

v1= (m1-m2/m1+m2)u1

0.91   =(0.21-m2/0.21+m2)1.4

0.91/1.4==(0.21-m2/0.21+m2)

from this eqn we can cal . m2 value

(c) speed of the center of mass

            Vcm = (m1v1+m2v2) / (m1+m2)                                 (   after impact )

by sub. we get ans.

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