Pls our usual way Straightforward formula no long process

Momentum is $p = mv$. Kinetic energy is $KE = \frac{1}{2}mv^2$.
We can express kinetic energy in terms of momentum:

We are given that $p_P = p_Q = p$ and $KE_Q > KE_P$.
Using the formula $KE = \frac{p^2}{2m}$:

Since $KE_Q > KE_P$, we have:

Since $p^2$ and 2 are positive, we can cancel them from the inequality:

This implies $m_P > m_Q$. So, Q weighs less than P.

Since $p_P = p_Q$, we have $m_P v_P = m_Q v_Q$.
Because $m_P > m_Q$, for the momentum to be equal, it must be that $v_Q > v_P$.
Therefore, Q is moving faster than P.

• Answer: (B) is moving faster than P

44.
The speed of the center of mass of a system remains constant if no external forces act on the system. A collision involves only internal forces, so the speed of the center of mass is conserved. We can calculate it using the initial conditions.

Mass of cart A: $m_A = 0.20 \, \text{kg}$
Initial velocity of cart A: $v_A = 3.0 \, \text{m/s}$
Mass of cart B: $m_B = 0.40 \, \text{kg}$
Initial velocity of cart B: $v_B = 2.0 \, \text{m/s}$ (traveling away from A, so in the same direction)

The speed of the center of mass $V_{CM}$ is given by:

• Answer: (C) $\boxed{\text{2.3 m/s}}$

45.
This problem involves the conservation of momentum during a collision.

Mass of ball A: $m_A = 200 \, \text{g} = 0.200 \, \text{kg}$
Mass of ball B: $m_B = 400 \, \text{g} = 0.400 \, \text{kg}$
Initial velocity of ball A: $v_{A,initial} = +0.3 \, \text{m/s}$ (assuming positive direction)
Initial velocity of ball B: $v_{B,initial} = ?$ (moving in opposite direction, so it will be negative)
Final velocity of ball A: $v_{A,final} = 0 \, \text{m/s}$ (comes to rest)
Final velocity of ball B: $v_{B,final} = 0 \, \text{m/s}$ (comes to rest)

The total momentum before the collision equals the total momentum after the collision:

The negative sign indicates that ball B was initially moving in the opposite direction to ball A, as stated in the problem.

• Answer: (C) $\boxed{\text{-0.15 m/s}}$