A block of mass $m$ slides on a frictionless surface and collides inelastically with a stationary magnet of mass $M$. If the two objects stick together after the collision, what is the speed $v$ of the combined block and magnet immediately after the collision?

• A. $\frac{mv{i}_1}{m}$
• B. $\frac{mv{i}_1}{M}$
• C. $\frac{Mv{i}_1}{m+M}$
• D. $\frac{mv{i}_1}{m+M}$

Answer: (D)

Verified step by step guidance

1. Identify the type of collision: Since the block and magnet stick together after the collision, this is a perfectly inelastic collision.
2. Write down the principle of conservation of momentum for the system: The total momentum before the collision equals the total momentum after the collision.
3. Express the initial momentum: The block of mass \(m\) has an initial velocity \(v_{i1}\), so its initial momentum is \(m \times v_{i1}\). The magnet of mass \(M\) is stationary, so its initial momentum is zero.
4. Express the final momentum: After the collision, the combined mass is \((m + M)\) moving with velocity \(v\). So the final momentum is \((m + M) \times v\).
5. Set up the conservation of momentum equation and solve for \(v\): \(m \times v_{i1} = (m + M) \times v\), which leads to \(v = \frac{m \times v_{i1}}{m + M}\).