Question

two cars with different masses travel at the same speed down a hill toward a stop sign. what will happen when both cars apply brakes at the same time to stop and identify and describe which of newton’s laws explains the result.

Explanation:

1. Situation Analysis: When the two cars (with different masses, \(m_1\) and \(m_2\), \(m_1

eq m_2\)) travel at the same speed (\(v\)) and apply brakes, we analyze their deceleration and stopping distance. Braking force (\(F\)) from the brakes (assuming same braking system, so force is related to friction, and for simplicity, assume the braking force magnitude per unit mass or the deceleration - causing force has a relationship with mass, but actually, the frictional force during braking is \(F_f=\mu N\), where \(N = mg\) (normal force, assuming horizontal braking on a flat part near the stop sign, or adjusted for the hill, but let's simplify). The deceleration \(a=\frac{F_{net}}{m}\), and if the braking force (from brakes + friction) is considered, for a car, when braking, the net force causing deceleration is \(F = ma\). If we assume the braking system provides a force proportional to mass (or the frictional force is \(\mu mg\), so \(F=\mu mg\), then \(a = \mu g\), which is independent of mass. Wait, but that's for friction - only braking. However, if the cars have different masses, and the brakes apply a force \(F\) (not proportional to mass), then \(a=\frac{F}{m}\). But in real - world, the braking force from the brake pads is related to the pressure and surface area, but the frictional force between tires and road is \(\mu N=\mu mg\). So the net deceleration \(a=\frac{\mu mg - F_{other}}{m}=\mu g-\frac{F_{other}}{m}\) (simplifying). But if we consider that the main decelerating force is friction, then \(a = \mu g\), independent of mass. So both cars, with same speed, same deceleration (if \(\mu\) is same, same road conditions), will take the same time to stop? Wait, no, let's use kinematics. The formula for stopping distance \(d=\frac{v^2}{2a}\), and time to stop \(t=\frac{v}{a}\). If \(a\) is same (because \(a = \mu g\), independent of mass, as long as \(\mu\) (coefficient of friction) and \(g\) (acceleration due to gravity) are same, and the braking force from the brakes doesn't change the deceleration in a mass - dependent way), then both cars will have the same deceleration. So with same initial speed \(v\), the time to stop \(t = \frac{v}{a}\) will be same, and stopping distance \(d=\frac{v^2}{2a}\) will be same? But wait, Newton's second law: \(F = ma\). The frictional force \(F_f=\mu N=\mu mg\). So the deceleration \(a=\frac{F_f}{m}=\mu g\), which is independent of mass. So regardless of mass, if the coefficient of friction and the braking - related forces (assuming friction is the main decelerating force) are same, both cars will decelerate at the same rate. So when they apply brakes at the same time, they will take the same time to stop and cover the same stopping distance (if initial speed is same and deceleration is same).

1. Newton's Law Explanation: Newton's second law (\(F = ma\)) explains this. The frictional force (decelerating force) \(F_f=\mu mg\). Rearranging \(F = ma\) to \(a=\frac{F}{m}\), substituting \(F = \mu mg\) gives \(a=\mu g\), which shows that acceleration (deceleration, in this case) is independent of mass. So even though the cars have different masses, the deceleration they experience (due to friction, assuming same road conditions) is the same. So with the same initial speed, they will stop in the same amount of time and cover the same distance (from the kinematic equations \(v = u+at\) (final velocity \(v = 0\), initial velocity \(u = v_{initial}\), so \(t=\frac{v_{initial}}{a}\)) and \(v^2 - u^2=2ad\) (so \(d=\frac{v_{initial}^2}{2a}\))).

Answer:

• What happens when brakes are applied: Both cars, despite having different masses, will decelerate at the same rate (assuming the same road conditions, so same coefficient of friction \(\mu\)) because the deceleration \(a=\mu g\) (from \(F = ma\) and \(F=\mu mg\)) is independent of mass. With the same initial speed \(v\), they will take the same time (\(t=\frac{v}{a}\)) to come to a stop and will cover the same stopping distance (\(d=\frac{v^2}{2a}\)).
• Newton's Law Explanation: Newton’s second law (\(F = ma\)) explains this. The frictional force (\(F_f=\mu mg\)) acting as the decelerating force gives \(a=\frac{F_f}{m}=\mu g\), showing deceleration is mass - independent. Thus, same - speed cars with different masses have the same deceleration, leading to same stopping time and distance.