Question

A small car of mass \(1000 .\) kg traveling at a speed of \(33.0 \mathrm{~m} / \mathrm{s}\) collides head on with a large car of mass \(3000 .\) kg traveling in the opposite direction at a speed of \(30.0 \mathrm{~m} / \mathrm{s}\). The two cars stick together. The duration of the collision is \(100 . \mathrm{ms}\). What acceleration (in \(g\) ) do the occupants of the small car experience? What acceleration (in \(g\) ) do the occupants of the large car experience?

Short answer

The occupants of the small car experience an acceleration of -48.22 g, and the occupants of the large car experience an acceleration of 16.07 g during the collision.

Step-by-step solution

Determining the initial and final velocity of each car

Let's first determine the initial velocity (u) and final velocity (v) of both small car (Sc) and large car (Lc). For Sc, u=33 m/s and mass (m) = 1000 kg; For Lc, u=-30 m/s (opposite direction) and mass (m) = 3000 kg;

Using the momentum conservation principle to find the final velocity of combined cars

The momentum conservation principle states that the total momentum before a collision is equal to the total momentum after the collision. So, we have: Initial total momentum = Final total momentum m1*u1 + m2*u2 = (m1 + m2)*v Plugging in the values: 1000*33 + 3000*(-30) = (1000 + 3000)*v Solve for 'v' (final velocity of both cars combined after collision): v = (-57000)/(4000) = -14.25 m/s

Calculating the change in velocity for each car

We will now find the change in velocity experienced by each car during the collision. For Sc, Δv = v_final - v_initial = (-14.25) - 33 = -47.25 m/s For Lc, Δv = v_final - v_initial = (-14.25) - (-30) = 15.75 m/s

Calculating the acceleration experienced by each car

From Newton's second law, Force (F) = mass (m) x acceleration (a). We also know that the duration of the collision is 100 ms. The change in momentum (Δp) is equal to the force (F) multiplied by time (t). So we can write: F = m * a = Δp So, a = Δp / m For Sc, a_Sc = Δv_Sc / t_Sc a_Sc = -47.25 / 0.100 Therefore, a_Sc = -472.5 m/s^2 For Lc, a_Lc = Δv_Lc / t_Lc a_Lc = 15.75 / 0.100 Therefore, a_Lc = 157.5 m/s^2

Converting accelerations to 'g' units

To convert the accelerations to 'g' units (where 1g = 9.81 m/s^2), we will divide each acceleration by 9.81. For the small car, a_Sc_g = a_Sc / 9.81 = -472.5 / 9.81 = -48.22 g For the large car, a_Lc_g = a_Lc / 9.81 = 157.5 / 9.81 = 16.07 g

Answer

The occupants of the small car experience an acceleration of -48.22 g, and the occupants of the large car experience an acceleration of 16.07 g.

Key concepts

Momentum Conservation

Momentum conservation is a fundamental principle in collision physics that states that when no external forces are acting, the total momentum of a system remains constant. In our exercise, we examine a collision between two cars of different masses moving at distinct speeds. According to this principle, the total momentum before the collision equals the total momentum after the collision. This is represented by the equation:

\[\begin{equation} m_1 \cdot u_1 + m_2 \cdot u_2 = (m_1 + m_2) \cdot v \end{equation}\]
Where:\

• \( m_1 \) and \( m_2 \) are the masses of the small and large cars,
• \( u_1 \) and \( u_2 \) are their initial velocities, and
• \( v \) is the final combined velocity after the cars stick together post-collision.

By applying the values, we can determine \( v \), and subsequently, the change in momentum, which is key to finding the force experienced during the collision. Understanding momentum conservation is crucial for solving collision problems, as it is preserved in all collisions irrespective of the object's size, shape, or mass.

Newton's Second Law

Newton’s second law of motion is intrinsically linked to the concepts of force and acceleration. It states that the acceleration of an object is directly proportional to the net force acting on it and is inversely proportional to its mass. This relationship can be expressed by the simple but powerful equation:

\[\begin{equation} F = m \cdot a \end{equation}\]
Where:

• \( F \) represents force,
• \( m \) is mass,
• \( a \) is acceleration.

The relevance of Newton's second law in collision scenarios, such as with our two cars, cannot be overstated. It allows the calculation of acceleration by re-arranging the equation to \( a = \frac{\Delta p}{m} \), using the change in momentum (\( \Delta p \)) over time, which we know from the conservation of momentum. By understanding this relationship, one can deduce the acceleration forces that occupants experience during collisions, as demonstrated in our exercise example.

Acceleration Due to Collision

Acceleration during a collision refers to how quickly an object changes its velocity as a result of an impact. This is a critical concept, as the rate of acceleration is a measure of the force experienced by the objects involved in a collision. The example given in our problem illustrates how to calculate the acceleration experienced by the cars and their occupants.

To derive the acceleration due to collision, one must consider the change in velocity (\( \Delta v \)) of the object and the duration of the collision (\( t \)). The formula used is:

\[ a = \frac{\Delta v}{t} \]
For the occupants of the small car and large car, the negative sign in acceleration for small car signifies that they were decelerated, or their speed decreased, more intensely than the large car. By converting the acceleration values into ‘g’ units, we are able to understand the severity of the impact relative to the force of gravity. Experiencing high accelerations (or decelerations) in terms of ‘g’ forces has tangible effects on the human body, which makes accurate calculations vital for safety considerations in vehicle design and collision analysis.