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(II) \(\mathrm{A} 920\) -kg sports car collides into the rear end of a 2300 -kg SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 2.8 \(\mathrm{m}\) before stopping. The police officer, estimating the coefficient of kinetic friction between tires and road to be 0.80 , calculates the speed of the sports car at impact. What was that speed?

Short Answer

Expert verified
The speed of the sports car at impact was approximately 12.8 m/s.

Step by step solution

01

Understand Conservation of Momentum

When two objects collide and lock together, we use the conservation of momentum to find the velocity just after the collision. The formula is: \[ m_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2) v_f \] where \( m_1 = 920 \) kg is the mass of the sports car, \( v_{1i} \) is its initial velocity, \( m_2 = 2300 \) kg is the mass of the SUV, \( v_{2i} = 0 \) since the SUV is at rest, and \( v_f \) is the final velocity of the combined mass just after the collision.
02

Calculate Frictional Force and Acceleration

The frictional force \( f_k \) is given by \( f_k = \mu_k (m_1 + m_2)g \) where \( \mu_k = 0.80 \) and \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity. This frictional force causes deceleration \( a \) which can be calculated using \( f_k = (m_1 + m_2)a \). Solve for \( a \): \[ a = \frac{\mu_k (m_1 + m_2)g}{m_1 + m_2} = \mu_k g = 0.80 \times 9.8 \] Calculate \( a \).
03

Apply Kinematic Equation

Since the cars stop after skidding 2.8 m, we can apply the kinematic equation: \[ v_f^2 = v_i^2 + 2ad \] Here, \( v_f = 0 \) (final velocity is 0), \( a \) is the acceleration calculated in Step 2 (negative because it's a deceleration), and \( d = 2.8 \text{ m} \). Rearrange to solve for \( v_i \): \[ v_i = \sqrt{-2ad} \]. Substitute \( a \) and \( d \).
04

Solve for Impulse Velocity

Use the calculated \( v_i \) from Step 3 as the velocity \( v_f \) in the momentum equation to solve for \( v_{1i} \): \[ 920 \cdot v_{1i} = (920 + 2300) \cdot v_i \] Solve for \( v_{1i} \).
05

Final Calculation

Finish the calculations based on the above formulas to find \( v_{1i} \). Use the values: \( a = 0.80 \times 9.8 = 7.84 \text{ m/s}^2 \), \( d = 2.8 \text{ m} \), and solve for the initial speed of the sports car before the collision.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coefficient of Kinetic Friction
The coefficient of kinetic friction, denoted as \( \mu_k \), is a crucial factor in understanding how objects slow down when sliding against each other. In this scenario, it represents the friction between the tires of the cars and the road surface. A police officer estimates this value to be 0.80 in the given problem, which is a relatively high friction level. This is significant in calculating the deceleration of the two cars after they collide and start skidding.
The kinetic frictional force can be calculated using the formula:
  • \[ f_k = \mu_k (m_1 + m_2)g \]
Here, \( g \) is the acceleration due to gravity, 9.8 \( \text{m/s}^2 \). The frictional force directly causes the deceleration, thus playing a vital role in predicting how cars will come to a stop after a collision.
Kinematic Equations
Kinematic equations are fundamental tools used to link different variables of motion such as velocity, acceleration, and displacement. In this exercise, we particularly leverage the following equation:
  • \[ v_f^2 = v_i^2 + 2ad \]
Given that the final velocity \( v_f \) is zero—since both vehicles come to a halt after skidding—this equation helps us find the initial velocity \( v_i \) just after the collision. The acceleration here is actually the deceleration due to friction, and it’s essential to use a negative sign for it, as it opposes the motion.
This understanding allows us to connect the dots from collision to complete stop over the distance \( d = 2.8 \text{ m} \). Rearranging the formula as \( v_i = \sqrt{-2ad} \) and substituting for \( a \) and \( d \), we find how fast the cars were moving right just after the collision.
Impulse and Momentum
Impulse and momentum are pivotal concepts in collision physics. Momentum is defined as the product of mass and velocity \( p = mv \). It serves as a key player when analyzing collisions, especially under the umbrella of its conservation law.
In collisions, total momentum before and after can be calculated using:
  • \[ m_1 v_{1i} + m_2 v_{2i} = (m_1 + m_2) v_f \]
Here, the impulse experienced by both vehicles during the collision translates into a momentum change, which is crucial for understanding post-impact motion. Impulse is essentially the integral of force over time and equals the change in momentum.
Using these principles, the problem calculates the velocity \( v_f \) right after the impact by substituting known masses and solving for the sports car's velocity before the collision, \( v_{1i} \).
Collision Physics
Collision physics often involves the concept of inelastic collisions, where colliding objects stick together, which is the case here—where the cars lock bumpers. In inelastic collisions, although kinetic energy isn’t conserved, momentum is. This makes the momentum conservation law a valuable tool for calculations.
In this exercise, the vehicles adhere, turning the scenario into a classic inelastic event. This framework allows application of combined mass (\( m_1 + m_2 \)) in subsequent steps, especially when calculating the displacement and deceleration post-impact.
Understanding collision physics in this context ensures the clear calculation of speed at which the sports car hit the SUV. Analyzing the motion from impact through deceleration to a stop effectively showcases how these specific physics principles apply in real-world scenarios.

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Most popular questions from this chapter

(I) The \(\mathrm{CM}\) of an empty \(1250-\mathrm{kg}\) car is \(2.50 \mathrm{~m}\) behind the front of the car. How far from the front of the car will the \(\mathrm{CM}\) be when two people sit in the front seat \(2.80 \mathrm{~m}\) from the front of the car, and three people sit in the back seat \(3.90 \mathrm{~m}\) from the front? Assume that each person has a mass of \(70.0 \mathrm{~kg}\).

(III) A 3.0-kg block slides along a frictionless tabletop at \(8.0 \mathrm{~m} / \mathrm{s}\) toward a second block (at rest) of mass \(4.5 \mathrm{~kg} . \mathrm{A}\) coil spring, which obeys Hooke's law and has spring constant \(k=850 \mathrm{~N} / \mathrm{m},\) is attached to the second block in such a way that it will be compressed when struck by the moving block, Fig. \(9-40 .\) ( \(a\) ) What will be the maximum compression of the spring? (b) What will be the final velocities of the blocks after the collision? \((c)\) Is the collision elastic? Ignore the mass of the spring.

(II) A mass \(m_{\mathrm{A}}=2.0 \mathrm{~kg},\) moving with velocity \(\overrightarrow{\mathbf{v}}_{\mathrm{A}}=\) \((4.0 \hat{\mathbf{i}}+5.0 \hat{\mathbf{j}}-2.0 \hat{\mathbf{k}}) \mathrm{m} / \mathrm{s},\) collides with mass \(m_{\mathrm{B}}=3.0 \mathrm{~kg}\) which is initially at rest. Immediately after the collision, mass \(m_{\mathrm{A}}\) is observed traveling at velocity \(\overrightarrow{\mathbf{v}}_{\mathrm{A}}^{\prime}=(-2.0 \hat{\mathbf{i}}+3.0 \hat{\mathbf{k}}) \mathrm{m} / \mathrm{s}\) Find the velocity of mass \(m_{\mathrm{B}}\) after the collision. Assume no outside force acts on the two masses during the collision.

(I) A 9150 -kg railroad car travels alone on a level frictionless track with a constant speed of \(15.0 \mathrm{~m} / \mathrm{s}\). A \(4350-\mathrm{kg}\) load, initially at rest, is dropped onto the car. What will be the car's new speed?

(I) A \(130-\mathrm{kg}\) tackler moving at \(2.5 \mathrm{~m} / \mathrm{s}\) meets head-on (and tackles) an \(82-\mathrm{kg}\) halfback moving at \(5.0 \mathrm{~m} / \mathrm{s}\). What will be their mutual speed immediately after the collision?

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