Problem 33 - College Physics

Question

A \(2 \mathrm{~kg}\) block is moving at a speed of \(10 \mathrm{~m} / \mathrm{s}\) and makes a perfectly elastic collision with a second block of mass \(M\), which is initially at rest. After the collision, the \(2 \mathrm{~kg}\) block bounces straight back at \(3 \mathrm{~m} / \mathrm{s}\). (a) Determine the mass \(M\) of the second block. (b) Determine the speed of the second block after the collision.

Step-by-Step Solution

Answer

The mass of the second block, \(M\), is 4.2 kg. The speed of the second block after the collision is \(5.1 \text{ m/s}\).

Step 1: Analyze Conservation of Momentum

In a perfectly elastic collision, the total momentum before the collision is equal to the total momentum after the collision. The initial momentum of the system is the momentum of the moving block, since the second block is initially at rest. The formula for momentum is given by the equation \( p = mv \). Initially, we have:\[p_{ ext{initial}} = (2 \, ext{kg})(10 \, ext{m/s}) = 20 \, ext{kg m/s}.\]

Step 2: Set Up the Equation for Conservation of Momentum

After the collision, the first block (2 kg) has reversed direction with speed 3 m/s, and the second block has some final speed \( v_2 \). Using conservation of momentum:\[20 \, ext{kg m/s} = (2 \, ext{kg})(-3 \, ext{m/s}) + (M \, ext{kg})(v_2) \]Simplifying gives:\[20 = -6 + Mv_2\]

Step 3: Analyze Conservation of Kinetic Energy

In a perfectly elastic collision, kinetic energy is also conserved. Therefore, the total kinetic energy before the collision equals the total kinetic energy after the collision. The initial kinetic energy is:\[KE_{ ext{initial}} = \frac{1}{2} (2 \, ext{kg})(10 \, ext{m/s})^2 = 100 \, ext{J}.\]

Step 4: Set Up the Equation for Conservation of Kinetic Energy

After the collision, the kinetic energies are as follows:For the 2 kg block: \[KE_{1, ext{final}} = \frac{1}{2}(2 \, ext{kg})(3 \, ext{m/s})^2 = 9 \, ext{J}.\]For the second block:\[KE_{2, ext{final}} = \frac{1}{2} M v_2^2.\]Using conservation of kinetic energy:\[100 = 9 + \frac{1}{2} M v_2^2\]

Step 5: Solve Simultaneous Equations for M and v2

We have two equations:1. \(20 = -6 + Mv_2\) 2. \(91 = \frac{1}{2} M v_2^2\) From equation 1:\[M v_2 = 26\]Substitute \(M v_2 = 26\) into equation 2:\[91 = \frac{1}{2}(26v_2/ M) \cdot v_2\]Rearrange to find \(v_2\):\[v_2^2 = 7 \, ext{m/s}\]Find \(v_2\):\[v_2 = 5.1 \, ext{m/s}\]Substitute back to find \(M\):\[91 = \frac{1}{2} M (5.1)^2\]\[M=4.2\, ext{kg}\]

Step 6: Verify Solution

Check both the momentum and energy conservation to ensure calculations are correct. Substitute values \(M = 4.2 \, ext{kg}\) and \(v_2 = 5.1 \, ext{m/s}\) back into the original equations to confirm both momentum and kinetic energy equations are satisfied.

Key Concepts

Conservation of Momentum

The conservation of momentum principle states that in a closed system, the total momentum before an event (like a collision) is equal to the total momentum after the event. Momentum itself is a product of mass and velocity, given by the equation \( p = mv \). In the context of our exercise, a moving block collides with another block at rest.To solve such a collision problem, you first calculate the total initial momentum, which here is only due to the moving block: \(p_{\text{initial}} = (2 \, \text{kg})(10 \, \text{m/s}) = 20 \, \text{kg m/s}\). Post-collision, the momentum of the two-block system must still equal this initial value. Balancing the equations helps us determine the unknowns, like the velocity and mass of the second block.

Conservation of Kinetic Energy

In a perfectly elastic collision, not only is momentum conserved, but kinetic energy is also conserved. Kinetic energy is given by the formula \( KE = \frac{1}{2}mv^2 \). Before the collision, only the first block has kinetic energy, calculated as \( KE_{\text{initial}} = \frac{1}{2} (2 \, \text{kg})(10 \, \text{m/s})^2 = 100 \, \text{J}\).After the collision, both blocks have kinetic energy. For the 2 kg block, you recompute its kinetic energy with its new velocity: \( KE_{1,\text{final}} = \frac{1}{2}(2 \, \text{kg})(3 \, \text{m/s})^2 = 9 \, \text{J}\). Conservation equations allow us to find the kinetic energy of the second block, helping us solve for unknown variables like the mass or velocity.

Collision Problems

Collision problems in physics require understanding and applying the laws of conservation of momentum and kinetic energy. In this exercise, a 2 kg block collides elastically with a stationary block, allowing us to use these laws to find unknowns like the mass and velocity of the second block. These types of problems can be tricky, often needing simultaneous equations for an accurate solution. By translating the physical scenario into mathematical equations, even a potentially confusing setup becomes manageable. Remember to always check your solution by putting the calculated values back into the original equations, ensuring both momentum and energy were indeed conserved.

Physics Problem Solving

Approaching physics problems systematically can greatly enhance understanding and accuracy. For collision problems, like the one we are tackling, start by clearly defining what you know and what you need to find.

• Identify the conservation laws applicable (momentum and kinetic energy for elastic collisions).
• Write down the initial conditions, such as masses and velocities.
• Set up equations representing conservation laws.
• Solve these equations, often simultaneously, for the unknowns.

With practice, this approach becomes a natural part of physics problem solving, allowing you to tackle increasingly complex scenarios efficiently.