In: Physics
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?
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