Question

A 1570 kg car skidding due north on a level frictionless icy road at 156 km/h collides with a 2245.1 kg car skidding due east at 120 km/h in such a way that the two cars stick together. At what angle (−180◦ ≤ θ ≤ +180◦ ) East of North do the two coupled cars skid off at?

Answer 1

Answer:

θ = 47.75º East of North.

Explanation:

• Assuming no external forces acting during the collision, total momentum must be conserved.
• Since momentum is a vector, if we project it along E-W and N-S axes, the momentum components along these axes must be conserved too.
• So, for the N-S axis, we can write the following equation:

       $p_{Northo} = p_{Northf} (1)$

• Since the car moving due east has no speed component along the N-S axis, the initial momentum along this axis is simply:

       $p_{Northo} = m_{1} * v_{1o} (2)$

      where m₁ = 1570 kg, v₁₀ = 156 km/h

• In order to work with the same units, we need to convert the speed in km/h to m/s, as follows:

       $v_{1o} = 156 km/h *\frac{1000m}{1 km}*\frac{1h}{3600s} = 43.3 m/s (3)$

• Replacing by the values in the left side of (1), we get:

       $p_{Northo} = m_{1} * v_{1o} = 1570 kg* 43.3 m/s = 67981 kg*m/s (4)$

• Since the collision is inelastic, both cars stick together, so we can write the right side of (1) as follows:

       $p_{Northf} = (m_{1} + m_{2})* v_{fNorth} = 3815.1 kg* v_{fNorth} (5)$

• From (4) and (5) , we can solve for VfNorth:

       $V_{fNorth} = \frac{67981kg*m/s}{3815.1kg} = 17.8 m/s (6)$

• We can repeat exactly the same process for the E-W axis:

       $p_{Easto} = p_{Eastf} (7)$

• Since the car moving due north has no speed component along the E-W axis, the initial momentum along this axis is simply:

       $p_{Easto} = m_{2} * v_{2o} (8)$

• As we did with v₁₀, we need to convert v₂₀ from km/h to m/s, as follows:

       $v_{2o} = 120 km/h *\frac{1000m}{1 km}*\frac{1h}{3600s} = 33.3 m/s (9)$

• Replacing by the values in the left side of (7), we get:

       $p_{Easto} = m_{2} * v_{2o} = 2245.1 kg* 33.3 m/s = 74762 kg*m/s (10)$

• Since the collision is inelastic, both cars stick together, so we can write the right side of (7) as follows:

       $p_{Eastf} = (m_{1} + m_{2})* v_{fEast} = 3815.1 kg* v_{fEast} (11)$

• From (10) and (11) , we can solve for VfEast:

       $V_{fEast} = \frac{74762 kg*m/s}{3815.1kg} = 19.6 m/s (12)$

• In order to find the angle East of North of the velocity vector, as we know the values of the horizontal and vertical components, we need just to apply a little bit of trigonometry, as follows:

       $tg \theta = \frac{V_{fEast}}{V_{fNorth} } = \frac{19.6}{17.8} = 1.1 (13)$

• The angle θ, East of North, is simply tg⁻¹ (1.1):

        θ = tg⁻¹ (1.1) = 47.75º E of N.