Question

1. a ball rolling on the floor with momentum $\boldsymbol{p_1}$ collides with a stationary ball and sets it in motion. the momentum of the first ball becomes $\boldsymbol{p_1}$, and that of the second becomes $\boldsymbol{p_2}$. compare the magnitudes of $\boldsymbol{p_1}$ and $\boldsymbol{p_2}$.a. momenta $\boldsymbol{p_1}$ and $\boldsymbol{p_2}$ are the same in magnitude.b. the sum of the magnitudes of $\boldsymbol{p_1}$ and $\boldsymbol{p_2}$ is zero.c. the magnitude of $\boldsymbol{p_1}$ is greater than that of $\boldsymbol{p_2}$.d. the magnitude of $\boldsymbol{p_2}$ is greater than that of $\boldsymbol{p_1}$.14. two cars are moving in the same direction. one car with momentum $\boldsymbol{p_1}$ collides with another, which has momentum $\boldsymbol{p_2}$. their momenta become $\boldsymbol{p_1}$ and $\boldsymbol{p_2}$ respectively. considering frictional losses, compare $(\boldsymbol{p_1}+\boldsymbol{p_2})$ with $(\boldsymbol{p_1}+\boldsymbol{p_2})$.a. the value of $(\boldsymbol{p_1}+\boldsymbol{p_2})$ is zero.b. the values of $(\boldsymbol{p_1}+\boldsymbol{p_2})$ and $(\boldsymbol{p_1}+\boldsymbol{p_2})$ are equal.c. the value of $(\boldsymbol{p_1}+\boldsymbol{p_2})$ will be greater than $(\boldsymbol{p_1}+\boldsymbol{p_2})$.d. the value of $(\boldsymbol{p_1}+\boldsymbol{p_2})$ will be greater than $(\boldsymbol{p_1}+\boldsymbol{p_2})$.

Explanation:

Question 13

Step1: Apply momentum conservation

According to the law of conservation of momentum \(p_1 = p_1'+p_2'\). Since \(p_1'\gt0\) (the first ball still has some momentum after collision), we can rewrite the equation as \(p_2'=p_1 - p_1'\).

Step2: Analyze the magnitude relationship

From \(p_2'=p_1 - p_1'\), we know that \(p_1=p_1'+p_2'\). So, \(p_1\gt p_2'\)

Step1: Consider the effect of frictional losses

In a real - world situation with frictional losses, the total momentum of the system is not conserved in an ideal sense. The frictional force is an external force that acts on the system of two cars.

Step2: Analyze the momentum change

The initial total momentum is \(p = p_1 + p_2\), and the final total momentum is \(p'=p_1'+p_2'\). Since frictional force acts in the opposite direction of the motion (tends to slow down the cars), \(p_1 + p_2\gt p_1'+p_2'\)

Answer:

C. The magnitude of \(p_1\) is greater than that of \(p_2'\)

Question 14