Question

In: Physics

Conservation of Momentum in Two Dimensions Ranking Task

The figures below show bird's-eye views of six automobile crashes an instant before they occur. The automobiles have different masses and incoming velocities as shown. After impact, the automobiles remain joined together and skid to rest in the direction shown by vfinal. Rank these crashes according to the angle θ , measured counterclockwise as shown, at which the wreckage initially skids.

 

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Solutions

Expert Solution

The required concepts to solve the problem are collision and momentum conservation.

First, use the momentum conservation to find a general equation for the angle made by the automobiles after the collision.

Then rank, from largest to smallest, the crashes according to the angle measured counterclockwise.

 

Velocity is a vector with both magnitude and direction. The direction of the vector depends on the angle made by the vector with respect to the horizontal or with respect to the vertical. If the angle is made with respect to the horizontal, then the x component of the vector is given by the cosθ component of the vector and the y component of the vector is given by the sinθ component of the vector.

According to the conservation of momentum, the momentum before the collision is equal to the momentum after the collision.

If the collision between two objects is considered, the conservation of momentum gives the relation,

m1u1 + m2u2 = m1v1 + m2v2

Here, m1 is the mass of the first object, m2 is the mass of the second object, u1 is the initial velocity of the first object, u2 is the initial velocity of the second object, v1 is the velocity after the collision of the first object and v2 is the final velocity after the collision of the second object.

Along the x direction, applying conservation of momentum,

(^u + °u) – eso many
o so mnya (^u + *u)=
osog ruyu+ soo uyu = nu

Here, mx is the mass of the car traveling in the x direction, my is the mass of the car travelling in the y direction, ux is the velocity of the car moving in the x direction, θ is the angle made by the combined mass with respect to the horizontal and vfinal is the velocity of the combined mass.

Along the direction, applying conservation of momentum,

mu, = m, Vinal sin 0+ m, Vinal sin 0
=(m +m,) sin e
Vam sin = (me +m,)

Here, uy is the velocity of the car moving in the y direction.

Finding the ratio of the components of the final velocity, we get

mu,
1 (m +m)
Vinal sin
final cos
mu
((m. +m)
tan (=y
mu

Thus, the equation for the angle made by the combined mass after collision is,

O = tan muy
mu

The equation for the angle made by the final velocity of the combined mass is,

O = tan muy
mu

For the first case, substitute 700kg for my, 1200kg for mx, 15 m/s for uy, and, 9 m/s for ux to find θ1.

0 = tan
700 kg)(15m/s)
|(1200kg)(9m/s))
= 44.2°

Here, θ1 is the angle in the first case.

For the second case, substitute 1000kg for my, 1000kg for mx, 10 m/s for uy, and 7 m/s for ux to find θ2.

0,= tan
((1000kg)(10m/s)
((1000 kg)(7m/s)
= 55°

Here, θ2 is the angle in the second case.

For the third case, substitute 700kg for my, 1000kg for mx, 7 m/s for uy, and 7 m/s for ux to find θ3.

0; = tan
((700 kg)(7m/s)
((700 kg)(7m/s))
= 45°

Here, θ3 is the angle in the third case.

For the fourth case, substitute 1000kg for my, 1000kg for mx, 7 m/s for uy, and 10 m/s for ux to find θ4.

Here, θ4 is the angle in the fourth case.

For the fifth case, substitute 1500kg for my, 1000kg for mx, 10 m/s for uy, and 15 m/s for ux to find θ5.

0x = tan
(1500 kg)(10m/s)
((1000 kg)(15m/s)
= 45°

Here, θ6 is the angle in the fifth case.

For the sixth case, substitute 1000kg for my, 1000kg for mx, 10 m/s for uy, and 10 m/s for ux to find θ6.

0 = tan
((1000kg)(10m/s)
|(1000 kg)(10m/s)
= 45°

Here, θ6 is the angle in the sixth case.


Ans:

The ranking of the crashes from largest to smallest is,

Vfinal
Vfinal
Vfinal
Vrina
Vfinal
m/s
15 m/s
10 m/s
9 m/s
10 m/s
1000 kg
700 kg
1000 kg
1000 kg
1200 kg
1000 kg
10 m/s 1
7 m/

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