Question

A 1500 kg car traveling east with a speed of 25 m/s collides at an intersection with a 2500 kg van traveling north at a speed of 20 m/s. The direction of wreckage after collision, assuming that the vehicles undergo a perfectly inelastic collision is:

• A. \(58.2^{\circ}\)
• B. \(47.2^{\circ}\)
• C. \(53.1^{\circ}\)
• D. \(50.6^{\circ}\)

Hint:

In 2D collision problems, always break the momentum into x and y components. Momentum is conserved independently in each direction. For a perfectly inelastic collision, the final velocity vector points in the same direction as the total initial momentum vector.

Answer 1

The Correct Option is C

Step 1: Define a coordinate system and calculate the initial momentum components. Let east be the positive x-direction and north be the positive y-direction. The collision is perfectly inelastic, so momentum is conserved. - Momentum of the car (x-direction): \(p_x = m_{car} v_{car} = (1500 \, \text{kg})(25 \, \text{m/s}) = 37500 \, \text{kg}\cdot\text{m/s}\). - Momentum of the van (y-direction): \(p_y = m_{van} v_{van} = (2500 \, \text{kg})(20 \, \text{m/s}) = 50000 \, \text{kg}\cdot\text{m/s}\). The initial total momentum is \(\vec{P}_{initial} = 37500\hat{i} + 50000\hat{j}\).
Step 2: Apply the law of conservation of momentum. For a perfectly inelastic collision, the vehicles stick together. The final momentum \(\vec{P}_{final}\) must be equal to the initial momentum. \[ \vec{P}_{final} = 37500\hat{i} + 50000\hat{j} \]
Step 3: Determine the direction of the final momentum vector. The direction of the wreckage is the direction of the final momentum vector. The angle \(\theta\) this vector makes with the positive x-axis (East) is given by: \[ \tan\theta = \frac{P_y}{P_x} = \frac{50000}{37500} \]
Step 4: Calculate the angle. \[ \tan\theta = \frac{500}{375} = \frac{20}{15} = \frac{4}{3} \approx 1.333 \] \[ \theta = \arctan\left(\frac{4}{3}\right) \approx 53.13^{\circ} \] The direction is approximately \(53.1^{\circ}\) (north of east).