Problem 92 - University Physics with Modern Physics

Question

A bullet with mass \(35.5 \mathrm{~g}\) is shot horizontally from a gun. The bullet embeds in a 5.90 -kg block of wood that is suspended by strings. The combined mass swings upward, gaining a height of \(12.85 \mathrm{~cm}\). What was the speed of the bullet as it left the gun? (Air resistance can be ignored here.)

Step-by-Step Solution

Answer

Answer: The initial velocity of the bullet was approximately 800.37 m/s.

Step 1: Determine the initial momentum of the bullet and the block

As the bullet embeds in the block of wood, we can assume that the momentum is conserved. The initial momentum of the bullet and the block is equal to their final momentum after the collision. Let the bullet's mass be \(m_1\) and its initial velocity be \(v_1\). The bullet's initial momentum is \(m_1v_1\). The block's mass is \(m_2\) and its initial velocity is zero \(v_2=0\). Thus, the initial momentum of the bullet and the block is: $$(m_1v_1)+m_2(0) = m_1v_1$$

Step 2: Determine the final momentum of the bullet and the block

After the bullet embeds in the block, they will have the same velocity due to being connected. Let their combined velocity be \(v\). The final momentum of the bullet and the block is given by: $$(m_1 + m_2)v$$

Step 3: Equate the initial and final momentum

Using the conservation of momentum, we equate the initial momentum and final momentum: $$m_1v_1 = (m_1 + m_2)v$$

Step 4: Use conservation of energy to find the height

At the highest point, the kinetic energy of the combined mass is converted into potential energy. We have: $$\frac{1}{2}(m_1 + m_2)v^2 = (m_1 + m_2)gh$$ Where \(g\) is the acceleration due to gravity and \(h\) is the height gained by the combined mass. We can solve for \(v^2\) as follows: $$v^2 = 2gh$$

Step 5: Calculate the initial velocity of the bullet

Using the expressions from Steps 3 and 4, we solve for \(v_1\): $$m_1v_1 = (m_1 + m_2)v = (m_1 + m_2)\sqrt{2gh}$$ $$v_1 = \frac{(m_1 + m_2)\sqrt{2gh}}{m_1}$$ Now, substitute the given values for the masses and height (\(m_1 = 35.5 \mathrm{~g}\), \(m_2 = 5.90 \mathrm{~kg}\), \(h = 12.85 \mathrm{~cm}\)) and the acceleration due to gravity (\(g \approx 9.81 \mathrm{~m/s^2}\)) into the equation and solve for \(v_1\): $$v_1 = \frac{(0.0355 + 5.9)\sqrt{2(9.81)(0.1285)}}{0.0355}$$ After calculating, we find the initial velocity of the bullet: $$v_1 \approx 800.37 \mathrm{~m/s}$$ The speed of the bullet as it left the gun was approximately \(800.37 \mathrm{~m/s}\).

Key Concepts

Conservation of Momentum

Momentum, the product of an object's mass and velocity, has a fundamental role in physics. Conservation of momentum is a principle stating that the total momentum of a closed system of objects (meaning no external forces are acting on it) remains constant over time. During a collision or interaction between two bodies, like a bullet and a block of wood, their total momentum before the collision is equal to their total momentum after the collision. This principle simplifies calculations in systems where colliding objects stick together, like in our exercise, allowing us to solve for unknown velocities.

Kinetic and Potential Energy

When we discuss motion and collisions, kinetic energy, the energy of an object in motion, is key. It's calculated using the formula \(\frac{1}{2}mv^2\), where \(m\) is mass and \(v\) is velocity. On the flip side, potential energy is the stored energy of an object due to its position or state. For example, when the bullet-block system in our exercise swings upwards, kinetic energy converts to gravitational potential energy, which is calculated with \(mgh\), where \(g\) is the acceleration due to gravity and \(h\) is the height. At the peak of its swing, all the system's kinetic energy has transformed into potential energy. This energy conservation allows us to relate velocity to the height the object reaches post-collision.

Projectile Motion in Physics

Projectile motion refers to the motion of an object thrown or projected into the air, subject only to the acceleration of gravity. The object is called a projectile, and its path is called its trajectory. The motion of a projectile can be broken into two components: horizontal and vertical. In ideal projectile motion, like the bullet in our textbook exercise, air resistance is typically ignored, leading to a constant horizontal velocity and a vertical acceleration due to gravity alone. As our problem's bullet-block system gained height, it essentially showcased projectile motion, where the vertical displacement was key to solving the problem.

Initial Velocity Calculation

Calculating the initial velocity is common in physics problems involving motion. The initial velocity is the velocity of an object before it undergoes acceleration or deceleration. To find it, we often use principles of conservation, like momentum and energy. In the context of our exercise, we sought the bullet's initial velocity. By equating the initial and final momentum and factoring in the energy conservation between kinetic and potential energy after the bullet embeds in the block, we find a formula involving masses, gravitational acceleration, and height. This formula lets us calculate the initial velocity before the impact, providing crucial insight into the bullet's speed as it left the gun.