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: Establish a coordinate system and compute initial momentum components. Define East as the positive x-direction and North as the positive y-direction. Since the collision is perfectly inelastic, 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 vector is \(\vec{P}_{initial} = 37500\hat{i} + 50000\hat{j}\).
Step 2: Apply the principle of momentum conservation. In a perfectly inelastic collision, the two vehicles merge. The final momentum, \(\vec{P}_{final}\), must equal the initial momentum. \[ \vec{P}_{final} = 37500\hat{i} + 50000\hat{j} \]
Step 3: Ascertain the direction of the final momentum vector. The direction of the wreckage is determined by the direction of the final momentum vector. The angle \(\theta\) this vector forms with the positive x-axis (East) is calculated as: \[ \tan\theta = \frac{P_y}{P_x} = \frac{50000}{37500} \]
Step 4: Compute 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.