Really hard Physics question. Requires much knowledge...?

Question

A vehicle (A) with a mass of 5kg and velocity of 0.5 m/s collides with a stationary vehicle (B) with a mass of 10kg. Determine the resulting velocities.

Answer 1

There are actually two ways to solve this problem. It just depends on what your assumptions are.

In the first solution, it can be assumed that the collision is perfectly INELASTIC, i.e., the two bodies are joined together after collision. Being so, the working equation is (applying the law of conservation of momentum)

MaVa + MbVb = (Ma + Mb)Vf

where

Ma = 5 kg (given)

Va = 0.5 m/sec (given)

Mb = 10 kg.

Vb = 0

Vf = final velocity of the Ma and Mb.

Substituting appropriate values,

(5)(0.5) + 10(0) = (5 + 10)Vf

Solving for Vf,

Vf = 2.5/15

Vf = 0.167 m/sec.

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For the second possible solution, it will be ASSUMED that the collision is perfectly ELASTIC hence the following working formulas are applicable.

Using the Law of Conservation of Momentum :

MaVa + MvVb = MaV1 + MbV2

where

V1 = final velocity of Ma after the collision

V2 = final velocity of Mb after the collision

and all the rest of the terms are as previously defined.

Substituting values,

5(0.5) + 10(0) = 5V1 + 10V2

5V1 + 10V2 = 2.5  --- Call this Equation 1

Using the Law of Conservation of Energy :

(1/2)MaVa^2 + (1/2)MbVb^2 = (1/2)MaV1^2 + (1/2)MbV2^2

Since "1/2" is a common factor, the above can be simplified to

MaVa^2 + MbVb^2 = MaV1^2 + MbV2^2

and substituting values,

5(0.5)^2 + 10(0)^2 = 5V1^2 + 10V2^2

5V1^2 + 10V2^2 = 1.25 --- Call this Equation 2

At this point, you have two equations with 2 unknowns. I trust that you know how to solve simultaneous equations with two unknowns.

Answer 2

Actually really easy physics question. It is a conservation of momentum problem. Assuming a perfectly inelastic collision (they stick together after the crash)...

(5 kg)(0.5 m/s) + (10 kg)(0 m/s) = (5 kg + 10 kg)(v_final)

You get 0.167 m/s in the direction the first car was going

Answer 3

Its quite simple the only thing you must understand is that momentum is always conseved whether its an elastic collision or an inelastic(where colliding bodies stick to each other after the collision) collision...

Now ..in this question you can assume that the collision between the two vehicles is a perfectly elastic(where after collion the vehicles dont 'stick' to each other!) and is a 'head -ON'(collision in one dimension).....

HOWEVER the fact is thaat it depends upon the situation ...

for exapmple consider this situation :-

a car with velocity 7m/s hits a bike ahead of it travelling in the same directing with a lesser velocity..there is a possibily that the bike may stick/get jammed into the front part of the car and then the system(car+bike) moves forward ...

but theree can be another situation as well where after the collision

the bike and the car dont stick to that extent and seperate -here its a partially elastic collision..

SO-in your case if assume a collision between ..say two cars and assume it to be a PERFECTLY ELASTIC COLLSION (which we generally do)  then-->

ANSWER IS-

M(A)=5Kg.

M(B)=10Kg.

V(a)=0.5 m/s

V(b)=0 m/s

during collision vehicle A transfers some of its energy to B conserving Kinetic energy as well(only in elastic coll.), so we have two equation:-

5*0.5=5V1+10V2  (consv. of momentum)

0.5 5*(0.5)^2=0.5 5*(V1)^2 +0.5 10*(V2)^2    (consv. of kinetic enegry)

so if you solve eqn. one square it and put in the value of of equation 2 --

then you get V1(of vehicle A)=  -1/6 m/s (minus sign indicates a velocity in direction opposite to the direction of velocity 0.5m/s with which it was travelling.))

V2= +1/3 m/s (in the direction of A's intial velocity vector)

so thats how you do it....