Question

In: Physics

Cart A traveling on a frictionless track at 6.0 m/s collides head-on with a 1.0 kg...

Cart A traveling on a frictionless track at 6.0 m/s collides head-on with a 1.0 kg cart B traveling in the opposite direction at 4.8 m/s. After the collision, cart A does not move. If the collision is perfectly elastic, a) What is the final velocity of cart B? b) what is the inertia of cart A? c) what is the kinetic energy of cart B after the collision?

Solutions

Expert Solution

We use conservation of moment, we estimate the amount of movement before and after the shock

. pi= m1 v1i + m2 v2i          v1i=6 m/s       v2i= -4.8 m/s

. pf = m1 v1f + m2 v2f        v1f=0

. pi = pf        m1 v1i + m2 v2i = m1 v1f + m2 v2f      m1( v1i-vif) = m2 ( v2f-v2i)

.   m1 ( 6 -0) = 1 (v2f + 4.8)

As the shock is elastic the kinetic energy is conserved

Ki=Kf                              ½ m1 v1i2 + ½ m2 v2i2 = ½   m2 v2f2   m1 62 + 1 4.82 = 1 v2f2

We write the system of equations and solve

6 m1 = v2f + 4.8

36 m1= v2f2 – 23.04

. v2f + 4.8 = v2f2 – 23.04                    v2f2 – v2f – 27.84 = 0    v2f =( -1 ±Ö (1 – 4 (-27.84)) ) / 2

. v2f = (-1 ±Ö 112.36 )/2 = (-1 ±10.6)/2

. v2f = 4.8 m/s

. v2f = -5.8 m/ s this solution implies that the car increases speed after the collision, which makes no sense, for this reason we reject this solution

The basket B final speed 4.8 m/s

.b) We are looking for the mass of the basket A m1

      6 m1 = v2f +4.8                       m1= (4.8+4.8)/6      m1= 1.6 Kg

     KA = ½ m1 v1i2       KA = ½ 1.6    62     KA= 28.8 J

.c) after the shock

KB = ½ m2 v2f2       KB = ½    1   4.82        KB= 11.52 J


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