Problem 1: Minimum Mass of Wooden Bar
- Newton’s Second Law: Force equals mass times acceleration ($F = ma$).
- Newton’s Third Law: Action and reaction forces are equal and opposite.
- Kinematics: Relationship between velocity, acceleration, and displacement ($v^2 = u^2 + 2as$).
Step 1: Calculate the deceleration of the bullet
The bullet enters the bar with speed $u = 100 \text{ m/s}$ and exits with speed $v = 80 \text{ m/s}$. The thickness of the bar is $d = 20 \text{ cm} = 0.2 \text{ m}$. Assuming uniform deceleration $a$, we use the kinematic equation:
$$v^2 = u^2 – 2ad$$Rearranging for $a$:
$$a = \frac{u^2 – v^2}{2d} = \frac{100^2 – 80^2}{2 \times 0.2} = \frac{10000 – 6400}{0.4} = \frac{3600}{0.4} = 9000 \text{ m/s}^2$$Step 2: Determine the force exerted by the bullet on the block
The resistive force $F$ exerted by the block on the bullet causes this deceleration. By Newton’s Second Law:
$$F = ma = 0.02 \text{ kg} \times 9000 \text{ m/s}^2 = 180 \text{ N}$$(Note: The mass of the bullet $m = 20 \text{ g} = 0.02 \text{ kg}$).
According to Newton’s Third Law, the bullet exerts an equal and opposite upward force $F = 180 \text{ N}$ on the wooden bar.
Step 3: Condition for the bar not to leave the tabletop
For the bar not to lift off the table, the upward force exerted by the bullet must be less than or equal to the gravitational force (weight) of the bar acting downwards.
$$F_{\text{upward}} \le Mg$$ $$180 \le M \times 10$$ $$M \ge 18 \text{ kg}$$Here, we neglected the weight of the bullet compared to the resistive force ($mg \approx 0.2 \text{ N} \ll 180 \text{ N}$), which is a standard approximation in such high-impact problems.