The impulse on the target will be the negative of that.From there you can write expressions for the momentum and velocity of the target post-collision.Clearly we want the bowling ball to have more mass than a pin, so that it can carry through to the pins behind the front pin(s).
The force that our dashboard or steering column exerts on us is going to equal our mass times our acceleration (as it constitutes our net horizontal force), and we are constrained to experience the same acceleration as our car.
So compare the accelerations of the two carts here.
Note that once again \(v_1=v,\;\;v_2=0\) is a solution (the incoming cart misses the target).
A clear application of this principle comes in bowling.
\begin momentum\;conservation: & mv 0 = mv_1 2mv_2 & \Rightarrow & v=v_1 2v_2 \\ elastic\;collision & \fracmv^2 0 = \fracmv_1^2 \frac2mv_2^2 & \Rightarrow & v^2=v_1^2 2v_2^2 \end \right\} \;\;\; \Rightarrow \;\;\; 2v_1=-v_2 \;\;\; \Rightarrow \;\;\; v_1=-\dfrac,\;\; v_2=\dfrac \] The lighter cart bounces off the heavier one at half the speed that the heavier one continues forward (or the incoming cart misses the target).
There is actually a clever way we could have solved this case more quickly by using the solution of the previous case and what we know about relative motion.A collision where the objects continue together with the same velocity after the collision (i.e.they remain stuck together), is often referred to as . It's not entirely clear that it'll be the same as it was in problem six, so we'll leave it unknown for now.Now you can get an expression of the impulse on the projectile, in terms of .More acceleration for our car means more acceleration for us, which means more force on us, which is bad.Lastly, we look at the lighter object bouncing off the heavier one: The math: \[ \left.The heavier cart goes from a speed v down to a speed of v/3, for a change of 2v/3.The lighter cart’s velocity changes from 0 to 4v/3 in the same period of time, which means it experiences twice the acceleration.Changing frames doesn't change the amount of internal energy created (it only changes the mechanical energy we see), so having the objects stick together results in the largest possible creation of internal energy.If we are told that a given collision is elastic (or at least can be approximated as such), then that gives us an additional condition that we can use to solve the problem. in each case, the diagram will show the experimental result, which we will then show mathematically using the combination of momentum and kinetic energy conservation.