Question

Consider two laboratory carts of different masses but identical kinetic energies and the three following statements. I. The one with the greatest mass has the greatest momentum II. The same impulse was required to accelerate each cart from rest III. Both can do the same amount of work as they come to a stop Which of the above statements would be correct?a. I onlyb. II onlyc. III onlyd. I and III only

Answer 1

Answer:

d) I and III only.

Explanation:

Let be $m_{1}$ and $m_{2}$ the masses of the two laboratory carts and let suppose that $m_{1} > m_{2}$. The expressions for each kinetic energy are, respectively:

$K = \frac{1}{2}\cdot m_{1}\cdot v_{1}^{2}$ and $K = \frac{1}{2}\cdot m_{2}\cdot v_{2}^{2}$.

After some algebraic manipulation, the following relation is constructed:

$\frac{m_{1}}{m_{2}} = \left(\frac{v_{2}}{v_{1}}\right)^{2}$

Since $\frac{m_{1}}{m_{2}} > 1$, then $\frac{v_{2}}{v_{1}} > 1$. That is to say, $v_{1} < v_{2}$.

The expressions for each linear momentum are, respectively:

$p_{1} = \frac{2\cdot K}{v_{1}} = m_{1}\cdot v_{1}$ and $p_{2} = \frac{2\cdot K}{v_{2}} = m_{2}\cdot v_{2}$

Since $v_{1} < v_{2}$, then $p_{1} > p_{2}$. Which proves that statement I is true.

According to the Impulse Theorem, the impulse needed by cart I is greater than impulse needed by cart II, which proves that statement II is false.

According to the Work-Energy Theorem, both carts need the same amount of work to stop them. Which proves that statement III is true.