Question

part 2 directions: look at the problem and the student work provided to determine and explain the errors made. then in the space provided show the correct work to reach the correct answer.

1. two carts roll towards each other on a frictionless track. cart has a mass of 850 kg and a velocity of 3m/s to the right. cart b has a mass of 300 kg and a velocity of 8m/s to the left. if the two carts join together after the collision, what is their final velocity?

student work:
$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$
$(850)(3) + (300)(-8) = 850v_{1f} + 300v_{2f}$
$2550 + (-2400) = 1150 + 2v_f$
$100 = 1150 + 2v_f$
$-1050 = 2v_f$
$v_f = -525$ m/s
error(s) explanation:

corrected work:

1. a 1000 kg car traveling initially with a velocity of 12m/s to the right crashes into a 500 kg truck moving to the left at 20 m/s. the velocity of the car after the collision is 7 m/s to the left. what is the velocity of the truck after the collision?

student work:
$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$
$(1000)(12) + (500)(20) = (1000)(7) + (500)v_{2f}$
$12000 + 10000 = 7000 + 500v_{2f}$
$22000 = 700 + 500v_{2f}$
$-700 -700$
$21300 = 500v_{2f}$
$\frac{21300}{500} = \frac{500v_{2f}}{500}$
$v_f = 42.6$
error(s) explanation:

corrected work:

Explanation:

Problem 4

Step1: Define sign convention

Let right = positive, left = negative. Since carts join, $v_{1f}=v_{2f}=v_f$.

Step2: Write conservation of momentum

$$m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v_f$$

Step3: Substitute values

$$(850)(3) + (300)(-8) = (850+300)v_f$$

Step4: Calculate left-hand side

$$2550 - 2400 = 150$$

Step5: Solve for $v_f$

$$v_f = \frac{150}{1150} = \frac{3}{23} \approx 0.13 \, \text{m/s}$$

Error(s) Explanation:

1. The student incorrectly simplified $850v_{1f}+300v_{2f}$ as $1150 + 2v_f$ (wrongly combined mass and velocity terms).
2. The student used an incorrect arithmetic step when subtracting 2550 - 2400, and incorrectly divided by 2 instead of the total mass 1150.

Step1: Define sign convention

Let right = positive, left = negative.

Step2: Write conservation of momentum

$$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$$

Step3: Substitute values

$$(1000)(12) + (500)(-20) = (1000)(-7) + 500v_{2f}$$

Step4: Calculate left-hand side

$$12000 - 10000 = 2000$$

Step5: Rearrange to solve for $v_{2f}$

$$2000 + 7000 = 500v_{2f}$$
$$9000 = 500v_{2f}$$
$$v_{2f} = \frac{9000}{500} = 18 \, \text{m/s}$$

Error(s) Explanation:

1. The student assigned the wrong sign to the car's final velocity: the car moves left, so $v_{1f} = -7$ m/s, not +7 m/s.
2. The student miscalculated the left-hand side sum and subsequent subtraction step, leading to an incorrect value.

Answer:

The final velocity is $\frac{3}{23} \approx 0.13$ m/s to the right.

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Problem 5