Problem 41 - College Physics

Question

A \(12.0-\mathrm{g}\) bullet is fired horizontally into a \(100-\mathrm{g}\) wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant \(150 \mathrm{~N} / \mathrm{m}\). The bullet becomes embedded in the block. If the bullet-block system compresses the spring by a maximum of \(80.0 \mathrm{~cm}\), what was the speed of the bullet at impact with the block?

Step-by-Step Solution

Answer

The final speed of the bullet at impact can be computed using the formula \(v_b=M*v_{ini}/m_b=sqrt{k*X^2/M}*sqrt{M/m_b}\) where all the values are given. Substitute all of them in to solve for \(v_b\).

Step 1: Determine the momentum before collision

The initial momentum of the bullet-block system is the mass of the bullet times the velocity of the bullet, as the block is initially at rest. Let's denote the mass of the bullet as \(m_b\), the velocity of the bullet as \(v_b\) and the mass of the block as \(m_{blk}\). The initial momentum \(P_{ini}\) is given by: \(P_{ini}=m_b*v_b\).

Step 2: Determine the momentum after collision

After the bullet is embedded in the block, they become a single object moving together. Therefore, the momentum of the bullet-block system after the collision is the mass of the system times its velocity. It's worth noting that this is also the initial velocity with which the combined system compresses the spring. This velocity \(v_{ini}\) is what we need to find. The combined mass \(M = m_{blk} + m_b\) and the final momentum \(P_{final}=M*v_{ini}\). According to the law of conservation of momentum in a closed system, the momentum before the collision should equal the momentum after the collision, that is, \(P_{ini}=P_{final}\). From here, we can express \(v_{ini}\) as: \(v_{ini}=P_{ini}/M=m_b*v_b/M\).

Step 3: Determine the speed of the bullet at impact

We now have everything we need to find the initial velocity of the bullet. Using the fact that the kinetic energy of the bullet-block system when the spring is at maximum compression is entirely converted from the initial kinetic energy, we can set up the equation \(1/2*M*v_{ini}^2=1/2*k*X^2\), where \(X\) is the maximum compression of the spring and \(k\) is the spring constant. From this, we can solve for \(v_b\), getting: \(v_b=M*v_{ini}/m_b=sqrt{k*X^2/M}*sqrt{M/m_b}\). Note that we have two square root terms here, but we know all of these values now and can simply substitute them in to solve for \(v_b\).

Key Concepts

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion. In our exercise, we see how kinetic energy plays a vital role in understanding the interactions within a bullet-block collision. The equation for kinetic energy is given by

• \( KE = \frac{1}{2} m v^2 \)

where \( m \) is the mass and \( v \) is the velocity of the object. When the bullet is fired into the block, the system's total kinetic energy is initially related solely to the bullet, as the block is at rest. During the collision, the bullet's kinetic energy gets transferred and shared with the block. This energy transformation leads to compressing the spring. Post-collision, the kinetic energy of the bullet-block system converts to potential energy stored in the spring when it compresses to its maximum extent.

Spring Constant

The spring constant, denoted as \( k \), is an essential parameter in this problem. It measures a spring's stiffness. The higher the spring constant, the more force is needed to compress or stretch it by a given distance. In the given exercise, the spring constant is crucial for calculating the bullet's velocity post-collision. Hooke’s Law helps relate the force exerted on the spring to its displacement:

• \( F = -kX \)

where \( F \) is the force exerted by the spring and \( X \) is the displacement. In our scenario, when the bullet-block system compresses the spring by \( 80 \) cm, the potential energy stored in the spring is

• \( PE = \frac{1}{2}kX^2 \).

This potential energy is equal to the initial kinetic energy of the bullet-block system, allowing us to find

• the speed of the bullet at impact.

Bullet-Block Collision

A bullet-block collision highlights the principles of the conservation of momentum and energy transformation. Before the impact, the bullet carries momentum while the block is at rest. Upon collision, the bullet embeds in the block, and they move together as a single unit. According to the

• law of conservation of momentum: \( m_b \times v_b = M \times v_{ini} \).

This fundamental law asserts that without external forces, the system’s total momentum is conserved during the collision. The bullet's initial momentum turns into the system's momentum after they unite. Thus, this law allows us to calculate the post-collision velocity of the combined bullet-block system.

Velocity Calculation

Calculating the bullet's velocity at impact involves combining

• the principles of conservation of momentum and energy.

Initially, the bullet had both momentum and kinetic energy. When the bullet embeds into the block, the resultant velocity post-collision can be calculated using momentum conservation, with the equation

• \( v_{ini} = \frac{m_b \times v_b}{M} \).

Subsequently, during maximum spring compression, all kinetic energy is converted to potential energy in the spring, expressed as

• \( \frac{1}{2}M \times v_{ini}^2 = \frac{1}{2}kX^2 \).

Solving these equations allows you to find

• the bullet's velocity:
• \( v_b = \sqrt{\frac{k \times X^2}{M}} \times \sqrt{\frac{M}{m_b}} \).

This methodical approach underscores how physics principles blend to unveil velocities in collision events.