Problem 70 - College Physics

Question

Two blocks collide on a frictionless surface. After the collision, the blocks stick together. Block A has a mass M and is initially moving to the right at speed v. Block B has a mass 2M and is initially at rest. System C is composed of both blocks. (a) Draw a force diagram for each block at an instant during the collision. (b) Rank the magnitudes of the horizontal forces in your diagram. Explain your reasoning. (c) Calculate the change in momentum of block A, block B, and system C. (d) Is kinetic energy conserved in this collision? Explain your answer. (This problem is courtesy of Edward F. Redish. For more such problems, visit http://www.physics.umd.edu/perg.)

Step-by-Step Solution

Answer

In terms of forces, block A and B experience equal magnitude forces but in opposite directions. The change in momentum for Block A, Block B, and system C can be calculated using the principle of conservation of momentum. The initial velocity of block B is 0, after the collision both blocks stick together moving with the same velocity \( v/3 \), implying the conservation of momentum. Finally, as this is an inelastic collision (where the blocks stick together after the collision), the kinetic energy is not conserved.

Step 1 - Drawing Force Diagrams

For each block, the forces acting on it during the collision can be represented with a diagram. For block A, there is an impulse force due to the collision with block B acting towards left and the force due to block A's movement acting to the right.For block B, there is an impulse force due to the collision on block A acting towards right as it was initially at rest.

Step 2 - Ranking The Forces

According to Newton's third law, 'for every action, there is an equal and opposite reaction', the magnitudes of the forces experienced by both blocks during the collision are equal but directions are opposite. Hence, the magnitudes of forces on block A and block B are the same.

Step 3 - Calculating Momentum Change

The change in momentum (\( \Delta p \)) can be computed as the difference between the final momentum (after collision) and the initial momentum (before collision). For block A, \( \Delta p_A = p_{A,final} - p_{A, initial} \) For block B, \( \Delta p_B = p_{B,final} - p_{B, initial} \) For system C (both blocks), \( \Delta p_C = p_{C,final} - p_{C, initial} \)We know that before the collision, block B was at rest so its initial momentum is 0. After the collision, both blocks stick together moving with the same velocity, say \( v1 \). The total momentum after the collision, \( p_{C,final} \), is the mass of the system (3M) multiplied by \( v1 \). From the conservation of momentum \( Mv = 3Mv1 \), we can compute \( v1 = v/3 \). And we substitute \( v1 \) into the momentum calculations to get the change in momentum for blocks A, B, and system C.

Step 4 - Conservation of Kinetic Energy

Kinetic energy is conserved in a collision if it is an elastic collision. In this case, the blocks stick together after the collision hence it is an inelastic collision. So, the kinetic energy is not conserved in this collision. The kinetic energy before the collision is \( 1/2 * M * v^2 \) while after the collision, the kinetic energy is \( 1/2 * 3M * (v/3)^2 = 1/6 * M * v^2 \) There is a loss in kinetic energy in this inelastic collision.

Key Concepts

Conservation of Momentum

Understanding the principle of conservation of momentum is essential when analyzing collisions. It states that the total momentum of a closed system remains constant if no external forces act on it. In the context of the exercise involving the two colliding blocks, despite the forces that the blocks exert on each other during the collision, the total momentum of the system, which includes both blocks, must remain unchanged from before to after the collision.

This principle is particularly useful because it applies even when collisions are not perfectly elastic. For instance, when block A with mass M and velocity v collides with block B, which is at rest, the momentum of system C (comprising both blocks) right after the collision is the same as the momentum of block A before the collision. Mathematically, it can be shown using the formula: \[ p_{C,final} = p_{A,initial} + p_{B,initial} \]Since block B is initially at rest, its momentum would be 0, and thus \[ p_{C,final} = p_{A,initial} \]where \( p_{A,initial} = Mv \). By sticking together after the collision, both blocks move with a new common velocity, highlighting that the momentum has been conserved, despite their inelastic interaction.

Kinetic Energy in Collisions

The concept of kinetic energy in collisions can often confuse students, but it's key to distinguishing between elastic and inelastic collisions. Kinetic energy, defined as the energy of an object due to its motion, is only conserved in elastic collisions. In inelastic collisions like the one described in the exercise, the kinetic energy is not conserved because some of it is transformed into other forms of energy such as heat, sound, or deformation.

Diving into our exercise, we can analyze the kinetic energy before and after the collision. Block A had kinetic energy just before the impact, given by the formula: \( KE_{A, initial} = \frac{1}{2}Mv^2 \). After the collision, the blocks stick together, forming a single object with a new velocity \( v1 \) and combined mass 3M. Its kinetic energy is then: \( KE_{C,final} = \frac{1}{2}(3M)v1^2 \). Comparing initial and final kinetic energies, we see a reduction: \( KE_{C,final} < KE_{A, initial} \), thus confirming that kinetic energy is not conserved in this inelastic collision.

Force Diagrams

Force diagrams, also known as free-body diagrams, are pivotal in visualizing and analyzing the forces acting upon objects, especially during collisions. Such diagrams help us to understand which forces cause which effects. For the exercise on hand, during the instant of collision, block A experiences a force to the left due to the impact with block B. This force is reflected in the force diagram as an arrow pointing leftwards. Conversely, block B experiences an equal and opposite force to the right, adhering to Newton's third law of motion, which states every action has an equal and opposite reaction.

A proper force diagram for each block will show these forces as arrows of equal length, representing their equal magnitude but opposite direction. This diagrammatic representation assists in comprehending the interactions between objects and is crucial for solving problems involving forces and momentum. Ultimately, understanding force diagrams allows us to correctly apply physical principles, such as the conservation of momentum, to real-world situations and theoretical problems alike.