Question

multiple choice question
a 375-kg object moving east at 22.0 m/s collides with a stationary 416-kg object. during the collision, the two objects stick together. what will be the velocity of the two attached objects immediately after the collision?

22.0 m/s east
10.4 m/s east
11.0 m/s west
10.4 m/s west

Explanation:

Step1: Identify the principle

This is a perfectly inelastic collision, so we use the law of conservation of momentum: $m_1v_1 + m_2v_2=(m_1 + m_2)v_f$.
Here, $m_1 = 375\space kg$, $v_1 = 22.0\space m/s$ (east), $m_2 = 416\space kg$, $v_2 = 0\space m/s$ (stationary).

Step2: Substitute values

Substitute into the formula: $(375\times22.0)+(416\times0)=(375 + 416)v_f$.
Calculate left side: $375\times22.0=8250$, $416\times0 = 0$, so left side is $8250$.
Right side: $375 + 416 = 791$, so $791v_f=8250$.

Step3: Solve for $v_f$

$v_f=\frac{8250}{791}\approx10.4\space m/s$.
Since the initial velocity of the first object is east, the final velocity is also east.

Answer:

B. 10.4 m/s east