Laws of Motion: Projectile Collision Calculation

[pdaniels77]

simple i know but i have just hit a blank.

a projectile of mass 250g traveling at 1200m/s impacts and sticks to a stationary 3500g object.
What will be the resulting velocity of the two combined objects...

Many thanks

 

[arildno]

What is the momentum of the two-body system prior to the collision?
What must it be just AFTER the collision?

 

[pdaniels77]

Thanks for getting back to me...
I am hoping that the resulting velocity will be in the region of 120m/s.
Regarding the momentum...the 3500g body is stationary and impact from the 250g body is in a straight line...think of 2 balls...with the smaller one impacting the larger one dead centre

 

[arildno]

"Regarding the momentum...the 3500g body is stationary and impact from the 250g body is in a straight line..."
The total of which equals the momentum the 250g body has on its own; agreed?

That quantity must be conserved after collision, for the composite body with mass 3500+250 g

If you do this correctly, since 3500=14*250, the new velocity will be 1/15 of the impact velocity just prior to the colision (80 m/s).

 

[pdaniels77]

ok...thanks...so inorder to obtain a resultant velocity of 120m/s of the combined mass, would it be better to increase the mass of the smaller object or increase its velocity or both

 

[arildno]

Well, if you look at the resultant velocity v=120, m_b mass of ball m_s mass of stationary object, v_0 velocity of ball prior to impact, you can write the equation to ensure 120m/s as:
\frac{1}{1+\frac{m_{s}}{m_{b}}}v_{0}=120
Thus, you have two parameters you can play about with independently, the mass ratio and the initial velocity. You have infinitely many solutions to your problem. Pick the one you like best.

 

[pdaniels77]

brilliant thank you

 

[pdaniels77]

also ...sorry to be a pain but how would you calculate the impact energy of 3500g body traveling at 120m/s

 

[arildno]

Well, you should be able to calculate the change in kinetic energy the station has absorbed by the collision by comparing its present kinetic energy level (remember to use m_s here as its mass!) relative to the initial level of energy.

But, you have YET more energy to account for!
In this type of inelastic collision, kinetic energy of the system of two bodies has become TRANSFERRED to other types of energy, such as to sound, heat, and the amount of energy needed to create permanent deformations in either, or both objects.

THAT amount of energy is simply the difference in the whole system's energy after the collision to the whole system's total kinetic energy before the collision.

 

[pdaniels77]

Hi SORRY TO BOTHER YOU AGAIN BUT U HAVE JUST BEEN PLAYING AROUND WIT THE FORUMLAE

 

[pdaniels77]

oops fat fingers ...i have just been playing around with the formulae you provided and i can not get to to provide the rightanswer...i know it is me being dim...but can you explain the figures to me...if i subsiture Vo with 1200 and ms 3500 and mb 250 i get 80...confused

 

[arildno]

oops fat fingers ...i have just been playing around with the formulae you provided and i can not get to to provide the rightanswer...i know it is me being dim...but can you explain the figures to me...if i subsiture Vo with 1200 and ms 3500 and mb 250 i get 80...confused

I told you the answer with your figures was 80 m/s. Right in post 4.