Question 359: Centre of Mass and Rotational Motion

Question

A bullet of mass $$m$$ moving with velocity $$v$$ is fired into a wooden block of mass $$M$$, If the bullet remains embedded in the block, the final velocity of the system is

Accepted Answer

When a bullet of mass $$m$$ is fired into a wooden block of mass $$M$$, and the bullet becomes embedded within the block, this scenario is governed by the law of conservation of momentum. According to the law of conservation of momentum, the total momentum of a system remains constant if no external forces act on the system. The formula for momentum is given by $$p = m \cdot v$$, where $$m$$ is the mass and $$v$$ is the velocity.
Before the collision, the momentum of the bullet is $$m \cdot v$$ (since it is moving) and the momentum of the block is $$0$$ (since it is stationary). Therefore, the total initial momentum of the system is:
$$p_{	ext{initial}} = m \cdot v + 0 = m \cdot v$$
After the collision, the bullet and the block move together as a single body with a combined mass of $$M + m$$ and a final velocity of $$v_f$$. Thus, the total final momentum of the system is:
$$p_{	ext{final}} = (M + m) \cdot v_f$$
By conserving momentum (setting the total initial momentum equal to the total final momentum), we can find the final velocity:
$$m \cdot v = (M + m) \cdot v_f$$
To solve for $$v_f$$, we can rearrange the equation:
$$v_f = \frac{m \cdot v}{M + m}$$
So, the final velocity of the bullet-block system is $$\frac{m \cdot v}{M + m}$$.
Therefore, the correct option is:
Option C: $$\frac{m v}{m+M}$$

Answer

$$\frac{m+M}{m}$$

Answer

$$\frac{M+m}{m v}$$

Answer

$$\frac{m v}{m+M}$$

Answer

$$\frac{v}{m(M+m)}$$

Centre of Mass and Rotational Motion

A bullet of mass $m moving with velocity v is fired into a wooden block of mass M$, If the bullet remains embedded in the block, the final velocity of the system is

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