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A BMW of mass \(2.0 \times 10^{3} \mathrm{kg}\) is traveling at $42 \mathrm{m} / \mathrm{s} .\( It approaches a \)1.0 \times 10^{3} \mathrm{kg}$ Volkswagen going \(25 \mathrm{m} / \mathrm{s}\) in the same direction and strikes it in the rear. Neither driver applies the brakes. Neglect the relatively small frictional forces on the cars due to the road and due to air resistance. (a) If the collision slows the BMW down to \(33 \mathrm{m} / \mathrm{s},\) what is the speed of the \(\mathrm{VW}\) after the collision? (b) During the collision, which car exerts a larger force on the other, or are the forces equal in magnitude? Explain.

Short Answer

Expert verified
Answer: The final velocity of the Volkswagen after the collision is 43 m/s. The forces exerted by each car during the collision are equal in magnitude due to Newton's third law.

Step by step solution

01

(a) Conservation of linear momentum

To determine the speed of the Volkswagen after the collision, we'll use the conservation of linear momentum principle. The total momentum before the collision must equal the total momentum after the collision. We can write this as: \(m_{1} v_{1i} + m_{2} v_{2i} = m_{1} v_{1f} + m_{2} v_{2f}\) where \(m_{1} = 2.0 \times 10^{3} \mathrm{kg}\) (mass of BMW) \(v_{1i} = 42 \mathrm{m/s}\) (initial velocity of BMW) \(m_{2} = 1.0 \times 10^{3} \mathrm{kg}\) (mass of Volkswagen) \(v_{2i} = 25 \mathrm{m/s}\) (initial velocity of Volkswagen) \(v_{1f} = 33 \mathrm{m/s}\) (final velocity of BMW) We need to solve for \(v_{2f}\) (the final velocity of the Volkswagen).
02

(a) Calculate the final velocity of the Volkswagen

Plugging the given values into the momentum conservation equation, we can rearrange the equation to solve for \(v_{2f}\): \(v_{2f} = \frac{m_{1} v_{1i} + m_{2} v_{2i} - m_{1} v_{1f}}{m_{2}}\) \(v_{2f} = \frac{(2.0 \times 10^{3} \mathrm{kg})(42 \mathrm{m/s}) + (1.0 \times 10^{3} \mathrm{kg})(25 \mathrm{m/s}) - (2.0 \times 10^{3} \mathrm{kg})(33 \mathrm{m/s})}{1.0 \times 10^{3} \mathrm{kg}}\) \(v_{2f} = \frac{84000 \mathrm{kg \thinspace m/s} + 25000 \mathrm{kg \thinspace m/s} - 66000 \mathrm{kg \thinspace m/s}}{1.0 \times 10^{3} \mathrm{kg}}\) \(v_{2f} = \frac{43000 \mathrm{kg \thinspace m/s}}{1.0 \times 10^{3} \mathrm{kg}}\) \(v_{2f} = 43 \mathrm{m/s}\) The final velocity of the Volkswagen after the collision is \(43 \mathrm{m/s}\).
03

(b) Compare the forces exerted by each car during the collision

According to Newton's third law, the force exerted by the BMW on the Volkswagen is equal in magnitude but opposite in direction to the force exerted by the Volkswagen on the BMW. Therefore, the forces are equal in magnitude.

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