Calculating Final Velocity in a Totally Inelastic Collision

In summary, the scenario involves a bigger, moving train (mbig) colliding with a smaller, stationary train (msmall). After the collision, the two trains stick together. The momentum conservation equation for this scenario can be written as m1v1f + m2v2f = m1v1i + m2v2i, where m1 and m2 represent the masses of the two trains. Since the smaller train is not moving initially, its initial velocity (v2i) is equal to 0. Solving for the final velocity (vf) of the two trains, we get vf = (mbigvbig i)/(mbig+msmall). This equation is correct.
  • #1
bmandrade
63
0

Homework Statement


A bigger, moving, vI]i[/I], collides with a smalled train that is not moving. Assume that after the collision the trains stick together. Assume the trains have different masses (mbig and msmall). Write doen the momentum conservation equation for the scenario. Solve for the final velocity of the two trains, vf


Homework Equations


conservation of momentum
m1v1f + m2v2f = m1v1i + m2v2i

The Attempt at a Solution


So i decided that m1 = mbig and m2=msmall
mbigvbig f + msmallvsmall f = mbigvbig i + msmallvsmall i

since vsmalli is at rest then = 0

mbigvbig f + msmallvsmall f = mbigvbig i

(mbig+msmall)*vbigsmall f = mbigvbig i

vbigsmall f= mbigvbig i/(mbig+msmall)


is this right??
 
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  • #2
Correct!
 
  • #3
bmandrade said:

Homework Statement


A bigger, moving, vI]i[/I], collides with a smalled train that is not moving. Assume that after the collision the trains stick together. Assume the trains have different masses (mbig and msmall). Write doen the momentum conservation equation for the scenario. Solve for the final velocity of the two trains, vf


Homework Equations


conservation of momentum
m1v1f + m2v2f = m1v1i + m2v2i

The Attempt at a Solution


So i decided that m1 = mbig and m2=msmall
mbigvbig f + msmallvsmall f = mbigvbig i + msmallvsmall i

since vsmalli is at rest then = 0

mbigvbig f + msmallvsmall f = mbigvbig i

(mbig+msmall)*vbigsmall f = mbigvbig i

vbigsmall f= mbigvbig i/(mbig+msmall)


is this right??

Looks fine by me.
 
  • #4
ok thanks
 

Related to Calculating Final Velocity in a Totally Inelastic Collision

1. What is a totally inelastic collision?

A totally inelastic collision is a type of collision in which two objects stick together and move as one after colliding. This means that the two objects have the same final velocity and momentum after the collision.

2. How does a totally inelastic collision differ from an elastic collision?

In an elastic collision, the two objects bounce off each other and maintain their individual shapes and velocities. In a totally inelastic collision, the two objects stick together and move with a common velocity.

3. What factors affect the outcome of a totally inelastic collision?

The mass and velocity of the objects involved in the collision are the main factors that affect the outcome. The larger the mass and velocity of the objects, the greater the momentum and kinetic energy transferred during the collision.

4. Is kinetic energy conserved in a totally inelastic collision?

No, in a totally inelastic collision, some kinetic energy is lost due to deformation and/or the generation of heat. However, momentum is conserved in all types of collisions.

5. What are some real-life examples of totally inelastic collisions?

Some examples of totally inelastic collisions include a car crash, a hammer hitting a nail, and a bullet hitting a target. In each of these scenarios, the objects involved stick together and move with a common velocity after the collision.

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