At the intersection of Texas Avenue and University Drive,
ablue, subcompact car with mass 950 kg traveling East on
universitycollides with a amrron pickup truck with mass 1900 kg
that istraveling north on Texas and ran a red light. The
twovehicles stick together as a result of the collision and, after
thecollision, the wreckage is sliding at 16.0 m/s in the
direction24.0 degrees east of north. Calculate the speed of
eachvehicle before the collision. the collision occurs during
aheavy rainstorm; you can ignore friction forces between
thevehicles and the wet road.

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fenixbm fenixbm · Answer 1 · 2019-09-21T01:26:50+02:00

Answer:

The initial speed of the truck is 21.93 m/s, and the initial speed of the car is 19.524 m/s

Explanation:

We can use conservation of momentum to find the initial velocities.

Taking the unit vector [tex]\hat{i}[/tex] pointing north and [tex]\hat{j}[/tex] pointing east, the final velocity will be

[tex] \vec{V}_f = 16.0 \frac{m}{s} \ ( \ cos(24.0 \°) \ , \ sin (24.0 \°) \ )[/tex]

[tex] \vec{V}_f = ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )[/tex]

The final linear momentum will be:

[tex]\vec{P}_f = (m_{car}+ m_{truck}) * V_f[/tex]

[tex]\vec{P}_f = (950 \ kg \ + 1900 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )[/tex]

[tex]\vec{P}_f = (2.850 \ kg \ ) * ( \ 14.617 \frac{m}{s} \ , \ 6.508 \frac{m}{s} \ )[/tex]

[tex]\vec{P}_f = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )[/tex]

As there are not external forces, the total linear momentum must be constant.

So:

[tex]\vec{P}_0= \vec{P}_f [/tex]

As initially the car is travelling east, and the truck is travelling north, the initial linear momentum must be

[tex]\vec{P}_0= ( m_{truck} * v_{truck}, m_{car}* v_{car} )[/tex]

so:

[tex]\vec{P}_0= \vec{P}_f[/tex]

[tex]( m_{truck} * v_{truck}, m_{car}* v_{car} ) = ( \ 41,658.45 \frac{ kg \ m}{s} \ , \ 18,547.8 \frac{kg \ m}{s} \ )[/tex]

so

[tex]\left \{ {{m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s} } \atop {m_{car} \ v_{car}=18,547.8 \frac{kg \ m}{s} }} \right.[/tex]

So, for the truck

[tex]m_{truck} \ v_{truck} = 41,658.45 \frac{ kg \ m}{s} [/tex]

[tex]1900 \ kg \ v_{truck} = 41,658.45 \frac{ kg \ m}{s} [/tex]

[tex] v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg} [/tex]

[tex]v_{truck} = \frac{41,658.45 \frac{ kg \ m}{s}}{1900 \ kg} [/tex]

[tex]v_{truck} = 21.93 \frac{m}{s}[/tex]

And, for the car

[tex]950 \ kg \ v_{car}=18,547.8 \frac{kg \ m}{s}[/tex]

[tex]v_{car}=\frac{18,547.8 \frac{kg \ m}{s}}{950 \ kg}[/tex]

[tex]v_{car}=19.524 \frac{m}{s}[/tex]