Two-dimensional perfectly inelastic collision between two vehicles

AI Thread Summary
The discussion centers on analyzing a perfectly inelastic collision between a truck and a car using conservation of momentum principles. The calculated final speed of the combined vehicles after the collision is 57.9 km/h, with a direction of 11.85° below the horizontal axis. The energy dissipated during the collision is approximately 4.77 × 10^5 J. Participants confirm the calculations and suggest using momentum as a vector for a more straightforward approach to determine final velocity and angle. The conversation emphasizes the importance of correctly applying momentum conservation equations to resolve discrepancies in results.
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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.
 
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I confirm your results.
 
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Chestermiller said:
I confirm your results.
Thank you for taking your time. I appreciate it!
 
Lone Wolf said:
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.
 
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