Question

a cart of mass of 420 g moving on a frictioless horizontal linear air track at an initial speed of 1.4 m/s undergoes an elastic collision with an initially stationary cart of unknown mass M. after the collision the initial cart moves with a speed pf 0.76 m/s. (a). what is the mass of the cart . b) what is the velocity of the second cart after the collision .

Answer 1

ELASTIC

m1 = 0.42 kg                          m2 = M

speeds before collision

u1 = 1.4 m/s                          u2 = 0 m/s

speeds after collision

v1 = 0.76 m/s                              v2 = ?

initial momentum before collision

Pi = m1*u1 + m2*u2

after collision final momentum

Pf = m1*v1 + m2*v2

from momentum conservation

total momentum is conserved

Pf = Pi

m1*u1 + m2*u2 = m1*v1 + m2*v2

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

from energy conservation

total kinetic energy before collision = total kinetic energy after collision

KEi = 0.5*m1*u1^2 + 0.5*m2*u2^2

KEf =   0.5*m1*v1^2 + 0.5*m2*v2^2

KEi = KEf

0.5*m1*u1^2 + 0.5*m2*u2^2 = 0.5*m1*v1^2 + 0.5*m2*v2^2

0.5*m1*(u1^2-v1^2) = 0.5*m2*(v2^2-u2^2).....(2)

solving 1&2

we get

u1 + v1 = v2 + u2

u1 - u2 = v2-v1

v2 = u1 - u2 + v1 ..........(3)

using 3 in 1

v1 = ((m1-m2)*u1 + (2*m2*u2))/(m1+m2)

v2 = ((m2-m1)*u2 + (2*m1*u1))/(m1+m2)

-----------------

0.76 = ((0.42-M)*1.4 + (2*M*0))/(0.42+M)

M = 0.124 kg = 124 g

v2 = ((124-420)*0 + (2*420*1.4))/(420+124)

v2 = 2.16 m/s