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In the study of mechanics, two important concepts often arise: power and collisions. Power tells us how quickly work is done or energy is transferred, which is crucial in understanding the performance of machines, engines, and even human activities like running. Collisions, on the other hand, describe what happens when two or more bodies interact suddenly, such as in vehicle crashes, sports, or particle physics.
Understanding power helps us analyze how efficiently energy is used or transferred in mechanical systems, while studying collisions allows us to predict the outcomes of impacts, including velocities and energy changes. These concepts are not only fundamental in physics but also have practical applications in engineering, safety design, and everyday life.
In this section, we will explore power and collisions from first principles, develop the necessary formulas, and solve problems step-by-step to build a solid understanding.
Power is defined as the rate at which work is done or energy is transferred. In simple terms, it tells us how fast energy is being used or converted from one form to another.
Mathematically, power \( P \) is given by the work done \( W \) divided by the time \( t \) taken:
For example, if a motor lifts a load by doing 1000 Joules of work in 10 seconds, the average power output is \( 100 \, \text{W} \).
Average power is the total work done divided by the total time taken, useful when the force or velocity changes over time.
Instantaneous power is the power at a specific instant, especially important when velocity or force varies continuously. It can be expressed as the dot product of force and velocity vectors:
For example, if a force of 10 N acts on an object moving at 2 m/s in the same direction, the instantaneous power is \( 10 \times 2 = 20 \, \text{W} \).
When two bodies collide, their velocities and energies change depending on the nature of the collision. Collisions are broadly classified into three types:
Understanding these types helps us apply the correct conservation laws when solving problems.
In all collisions, the total momentum of the system before and after collision remains the same, provided no external forces act on the system. Momentum is the product of mass and velocity.
Mathematically, for two bodies with masses \( m_1 \) and \( m_2 \), initial velocities \( u_1 \) and \( u_2 \), and final velocities \( v_1 \) and \( v_2 \), the conservation of momentum states:
The coefficient of restitution \( e \) measures how elastic a collision is. It is defined as the ratio of the relative velocity of separation to the relative velocity of approach along the line of impact.
Its value ranges from 0 to 1:
Mathematically,
Work done by forces during collisions changes the kinetic energy of the bodies involved. In elastic collisions, kinetic energy before and after collision remains the same. In inelastic collisions, some kinetic energy is converted into other forms like heat or sound.
The kinetic energy \( KE \) of a body of mass \( m \) moving at velocity \( v \) is:
Work done \( W \) by a force \( \vec{F} \) moving an object through displacement \( \vec{d} \) at angle \( \theta \) is:
When solving problems involving power and collisions, follow these steps:
Step 1: Calculate the force required to lift the load at constant speed. Since speed is constant, force equals weight:
\( F = m g = 50 \times 9.8 = 490 \, \text{N} \)
Step 2: Calculate power using \( P = F \times v \):
\( P = 490 \times 0.5 = 245 \, \text{W} \)
Answer: The motor delivers 245 Watts of power.
Step 1: Since masses are equal and collision is elastic, velocities are exchanged.
Initial velocities: \( u_1 = 3 \, m/s \), \( u_2 = -2 \, m/s \) (opposite directions)
Step 2: After collision, \( v_1 = u_2 = -2 \, m/s \), \( v_2 = u_1 = 3 \, m/s \).
Answer: The first ball moves at -2 m/s, and the second ball moves at 3 m/s after collision.
Step 1: Use conservation of momentum:
\( m_1 u_1 + m_2 u_2 = (m_1 + m_2) v \)
\( 3 \times 4 + 2 \times (-3) = 5 \times v \)
\( 12 - 6 = 5 v \Rightarrow v = \frac{6}{5} = 1.2 \, m/s \)
Step 2: Calculate initial kinetic energy:
\( KE_i = \frac{1}{2} \times 3 \times 4^2 + \frac{1}{2} \times 2 \times 3^2 = 24 + 9 = 33 \, J \)
Step 3: Calculate final kinetic energy:
\( KE_f = \frac{1}{2} \times 5 \times (1.2)^2 = 3.6 \, J \)
Step 4: Kinetic energy lost:
\( \Delta KE = KE_i - KE_f = 33 - 3.6 = 29.4 \, J \)
Answer: The common velocity after collision is 1.2 m/s, and 29.4 Joules of kinetic energy is lost.
Step 1: Write conservation of momentum equation:
\( 4 \times 5 + 6 \times (-3) = 4 v_1 + 6 v_2 \)
\( 20 - 18 = 4 v_1 + 6 v_2 \Rightarrow 2 = 4 v_1 + 6 v_2 \) - (1)
Step 2: Use coefficient of restitution formula:
\( e = \frac{v_2 - v_1}{u_1 - u_2} = 0.8 \)
\( 0.8 = \frac{v_2 - v_1}{5 - (-3)} = \frac{v_2 - v_1}{8} \Rightarrow v_2 - v_1 = 6.4 \) - (2)
Step 3: Solve equations (1) and (2):
From (2): \( v_2 = v_1 + 6.4 \)
Substitute in (1):
\( 2 = 4 v_1 + 6 (v_1 + 6.4) = 4 v_1 + 6 v_1 + 38.4 = 10 v_1 + 38.4 \)
\( 10 v_1 = 2 - 38.4 = -36.4 \Rightarrow v_1 = -3.64 \, m/s \)
\( v_2 = -3.64 + 6.4 = 2.76 \, m/s \)
Answer: After collision, the first body moves at -3.64 m/s and the second at 2.76 m/s.
Step 1: Calculate power using \( P = F \times v \):
\( P = 50 \times 5 = 250 \, \text{W} \)
Answer: The power output of the person is 250 Watts.
When to use: When solving collision problems to decide which conservation laws to apply.
When to use: In problems involving coefficient of restitution to avoid lengthy algebra.
When to use: When instantaneous power needs to be found in mechanical systems.
When to use: When bodies stick together or deform during collision.
When to use: Always, especially in metric system based problems.
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