Momentum, impulse and force (mathematical)

1. Aug 21, 2012

hms.tech

1. The problem statement, all variables and given/known data
See the attachment
Once you read the question (attached) you'll notice that there are two different parts of the question, the first one asks to find the combined velocity of the system after collision while the 2nd one uses the result of the first part and states to find the force.

2. Relevant equations

F=ma
Acceleration due to gravity = g
(for constant acceleration) average velocity= (v+u)/2
Impulse, I = Ft (for a constant force)

3. The attempt at a solution

I solved the first part by finding the velocity of the combined system to be "x"
x = mv/(M+m)

but i am unable to solve the 2nd part of the question.

Here is what i did :

Momentum(initial) = mv
Momentum(final) = x(M+m)
Force (with which the ground resists penetration) = (m+M)x/t

As the force is constant thus the acceleration would also be constant, thus :
average velocity = [v(initial) +v(final)]/2
v(initial) = v one the other hand v(final) = 0 [because the system comes to rest]
thus average velocity = v/2
Now, s=vt
thus h=vt/2

by solving for "t" and then substituting "t" in the Force(with which the ground resists penetration) equation we get the final result of the force.

The result found by my method is wrong, instead the solution to this question involves the use of Weight of the combined system which is downward thus acting opposite to the Force(with which the ground resists penetration).
How can we take "Weight = (M+m)g" into account in this question, how do i include "g" in my answer to make it match with the correct solution ?

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Last edited: Aug 21, 2012
2. Aug 21, 2012

Aimless

Have you considered trying to solve for the force using work and energy instead?

Edit: As an alternative, try drawing a force diagram. The force you will obtain by solving for the acceleration is the net force; how do gravity and the resistance of the ground combine to give net force?

Last edited: Aug 21, 2012
3. Aug 21, 2012

hms.tech

That is exactly what i was thinking :

F(net) is the the one i just evaluated
also F(net) = F(resistance of ground) + Weight

So, by vector addition, F(net)-weight = F(resistance of ground)

the answer of F(resistance of Ground) i get is :

(M+m)[x^2 +2gh]/2h

but this is wrong, however it looks quite similar in form to the correct solution which is :

{ (mv)^2 + 2gh(M+m)^2 }/{ 2h(M+m) }

4. Aug 21, 2012

Aimless

Try plugging in the value you calculated for x into (M+m)[x^2 +2gh]/2h and see what happens.

5. Aug 21, 2012

hms.tech

yeah, that does it ... Thank you :)