Help on physics collision problem finding speed & kinetic energy given weights and initial speed?

Question

A model-train car of mass 260 g traveling with a speed of 0.50 m/s links up with another car of mass 430 g that is initially at rest.

(a) What is the speed of the cars immediately after they have linked together?

_____ m/s

(b) Find the initial kinetic energy.

_____ mJ

(c) Find the final kinetic energy.

_____ mJ

6 8.P.099.

Answer 1

mV = Mv

.260*.5 = .690V

V = .188m/s

E = (mv^2)/2

E = .260*.5*.5/2 = 0.0325j

E = (mv^2)/2

E = .690* .188* .188/2

E = 0.01219j

Hope this Helps

Answer 2

a) Conservation of energy:

m1*v1+m2*v2=(m1+m2)*V

260*.5+430*0=690*V

V=0.18m/s

b) k1=.5*m1*v1^2

c) Kf=.5*(m1+m2)*V^2

Answer 3

This is all to do with the conservation of momentum, which states that the total momentum of a closed system must remain constant, unless acted upon by an external force.

In this case there is no external force, so we can use this rule.

Momentum (from now on known as p) is found by the following equation....

Momentum(kg m/s) = Mass (kg) x Velocity (m/s)

So to find the answer to a) we state the rule of conservation of momentum...

        

             Total p before collision=Total p after collision

(0.26kg x 0.5m/s) + (0.43g x 0m/s)=(0.26kg + 0.43kg) x?m/s

          

                 0.13kg m/s + 0kg m/s =    0.69kg x ?m/s

A quick rearangment of this equation gives....

?m/s = 0.13kg m/s / 0.69kg

         = 0.19m/s

b) Kinetic energy= 1/2 mass x Velocity^2

            (train)       = 0.5 x 0.26kg x (0.5m/s)^2

                             = 0.0325J......32.5mJ

The car has no kinetic energy as it has zero velocity

c)Same equation as b)

          = 0.5 x 0.69kg x (0.19m/s)^2

          = 12.5mJ

Answer 4

Use conservation of momentum

M1V1 + M2V2 = (M1 + M2)V3

where

M1 = 0.260 kg.

V1 = 0.50 m/sec.

M2 = 0.430 kg.

V2 = 0

V3 = velocity of train/car after getting linked

Substituting values,

(0.26)(0.50) + (.043)(0) = (0.26 + 0.43)V3

Solving for V3,

V3 = 0.19 m/sec.

**********************************

Initial kinetic energy = 1/2(M1)(Vi^2)

Initial kinetic energy = (1/2)(0.260)(0.50)^2 = 0.0325 joules = 32.5 mJ

**********************************

Final kinetic energy = (1/2)(M1 + M2)V3^2

Final kinetic energy = (1/2)(0.260 + 0.430)(0.19)^2

Final kinetic energy = 0.0124 joules = 12.4 mJ