Two-dimensional perfectly inelastic collision between two vehicles

[Lone Wolf]

Homework Statement

A truck with mass 7.5 ton moving at a speed of 65 km/h in the west-east direction, collides with a car of mass 1100 kg moving from north to south at a speed of 93 km/h. After the collision, both vehicles move together.
a) What is the speed and direction of motion of the two vehicles after the collision?
b) How much energy is dissipated in the collision?

Relevant Equations

Conservation of linear momentum equation: m1v1 + m2v2 = m1v1f + m2v2f
Kinetic energy equation: 1/2 mv²

a) Let m be the vehicle's mass, M the truck's mass, vt the truck's speed, vc the car's speed, vf the final speed, θ the angle both vehicles make with the horizontal axis (west-east direction) after the collision.
Conservation of linear momentum:
In the x direction: M vt = (m + M) vf cos(θ)
In the y direction (I considered the positive direction from south to north): - m vc = (m + M) vf sin(θ)
Solving the system for θ:
θ= arctan( (-m vc)/(M vt) ) = ( (-1.1*93)/(7.5*65) ) = -11.85°
Solving for vf:
vf = (M vt)/((m + M) cos (θ)) = (7.5 * 65)/( (7.5 + 1.1)*cos(-11.85°) ) = 57.9 km/h
So the speed of the vehicles would be 57.9 km/h with the direction 11.85° below the horizontal axis.

b) ΔK = Kf - Ki = [(7.5 + 1.1)*1e3 * (57.9e3/3600)² - 7.5e3 * (65e3/3600)² - 1.1e3 * (95e3/3600)²]/2 = - 4,77 × 10^5 J
So 4,77 × 10^5 J were dissipated with the collision

The solutions given are:
a) 48.9 km/h, 55°
b) 0.8 × 10^6 J

I've tried doing this problem twice now and I always get the same result. Please help me find my error. Thanks.

 

[Chestermiller]

I confirm your results.

 

[Lone Wolf]

I confirm your results.

Thank you for taking your time. I appreciate it!

 

[PeroK]

Problem Statement: A truck with mass 7.5 ton moving at a speed of 65 km/h in the west-east direction, collides with a car of mass 1100 kg moving from north to south at a speed of 93 km/h. After the collision, both vehicles move together.
a) What is the speed and direction of motion of the two vehicles after the collision?
b) How much energy is dissipated in the collision?
Relevant Equations: Conservation of linear momentum equation: m1v1 + m2v2 = m1v1f + m2v2f
Kinetic energy equation: 1/2 mv²

a) Let m be the vehicle's mass, M the truck's mass, vt the truck's speed, vc the car's speed, vf the final speed, θ the angle both vehicles make with the horizontal axis (west-east direction) after the collision.
Conservation of linear momentum:
In the x direction: M vt = (m + M) vf cos(θ)
In the y direction (I considered the positive direction from south to north): - m vc = (m + M) vf sin(θ)
Solving the system for θ:
θ= arctan( (-m vc)/(M vt) ) = ( (-1.1*93)/(7.5*65) ) = -11.85°
Solving for vf:
vf = (M vt)/((m + M) cos (θ)) = (7.5 * 65)/( (7.5 + 1.1)*cos(-11.85°) ) = 57.9 km/h
So the speed of the vehicles would be 57.9 km/h with the direction 11.85° below the horizontal axis.

b) ΔK = Kf - Ki = [(7.5 + 1.1)*1e3 * (57.9e3/3600)² - 7.5e3 * (65e3/3600)² - 1.1e3 * (95e3/3600)²]/2 = - 4,77 × 10^5 J
So 4,77 × 10^5 J were dissipated with the collision

The solutions given are:
a) 48.9 km/h, 55°
b) 0.8 × 10^6 J

I've tried doing this problem twice now and I always get the same result. Please help me find my error. Thanks.

Here's an interesting idea using momentum as a vector. Conservation of momentum means it's the same before and after the collision. So, you could have calculated the total momentum of the system before the collision. Let's use $\vec{p}$ for momentum.

$(M+m)\vec{v_f} = \vec{p_f} = \vec{p_i} = (Mv_t, mv_c)$

So, all the information about the magnitude and direction of the final momentum is already all there in the initial velocities and masses of the vehicles.

To get the final velocity, you can just divide this by the combined mass $M+m$. And you can just read off the tangent of the angle $\theta$, which is also the same before and after.