Question

question 3 1 pts whether a truck comes to a stop by crashing into a haystack or a brick wall, the stopping force is greater with the brick wall. both the same greater with the haystack. question 4 1 pts two vehicles with equal magnitudes of momentum traveling at right angles to each other undergo an inelastic collision. the magnitude of momentum for the combined wreck is less than the magnitude of momentum of either car before collision. the same as the magnitude of momentum of either car before collision. greater than the magnitude of momentum of either car before collision. none of the above

Explanation:

Question 3

Step1: Recall impulse - momentum theorem

Impulse $J = F\Delta t=\Delta p$. The change in momentum $\Delta p$ is the same in both cases (truck goes from moving to stopped). The time $\Delta t$ is longer when hitting a hay - stack compared to a brick wall.

Step2: Analyze the force

Since $F=\frac{\Delta p}{\Delta t}$, with $\Delta p$ constant, a smaller $\Delta t$ (brick wall) gives a larger $F$. So the stopping force is greater with the brick wall.

Question 4

Step1: Represent momenta as vectors

Let the momentum of one vehicle be $\vec{p}_1$ and the other be $\vec{p}_2$. Since they are at right - angles, if $|\vec{p}_1| = |\vec{p}_2|=p$.

Step2: Use vector addition for in - elastic collision

By the law of conservation of momentum, the magnitude of the combined momentum $|\vec{P}|$ of the two vehicles after in - elastic collision is given by $|\vec{P}|=\sqrt{\vec{p}_1^{2}+\vec{p}_2^{2}}$ (vector addition of perpendicular vectors). Since $|\vec{p}_1| = |\vec{p}_2| = p$, $|\vec{P}|=\sqrt{p^{2}+p^{2}}=\sqrt{2}p>p$. So the magnitude of momentum for the combined wreck is greater than the magnitude of momentum of either car before collision.

Answer:

Question 3: greater with the brick wall.
Question 4: greater than the magnitude of momentum of either car before collision.