Question

A cart of mass 340 g moving on a frictionless linear air track at an initial speed of 1.2 m/s undergoes an elastic collision with an initially stationary cart of unknown mass. After the collision the first cart continues in its original direction at .66m/s.
a) what is the mass of the second cart?
b)What is its speed after impact?
c)What is the speed of the two-cart center of mass?

I know that in an elastic collsion momentum and energy is conserved. What would be the equation i would use for this problem? I can't figure out the equations for a) and b) but I think for
c) would I use V= (m_1* v_1 + m_2 *v_2)/(m1+m2)

Answer 1

The equation for the first part is

V_1Final = [(m_1-m_2)/(m_1+m_2)]*V_1Origional

I used conservation of momentum for the second part.

Answer 2

Given:
M1 = ?, V1 = 0.
M2 = 0.34kg, V2 = 1.2 m/s.
V3 = Velocity of M1 after the collision.
V4 = 0.66 m/s = Velocity of M2 after the collision.

a. M1*V1 + M2*V2 = M1*V3 + M2*V4.
M1*0 + 0.34*1.2 = M1*V3 + 0.34*0.66,
0 + 0.408 = M1*V3 + 0.2244,
Eq1: M1*V3 = 0.1836.

Conservation of KE Eq:
V3 = (V1(M1-M2) + 2M2*V2)/(M1+M).
V3 = (0*(M1-M2) + 0.816)/( M1+M2),
V3 = 0.816/(M1+0.34).

In Eq1, replace V3 with 0.816/(M1+0.34):
M1*0.816/(M1+0.34) = 0.184.
0.816M1 = 0.184M1 + 0.0626,
0.632M1 = 0.0626,
M1 = 0.10 kg.

Answer 3

b. V3 = 0.816/(0.10+0.34) = 1.85 m/s. = Velocity of M1 after the collision.