Based on the information provided, what is the mass of the truck?As a 1250 kg car was passing through an intersection at 8.12 m/s [0.0 degrees] it was hit broadside by a pickup truck...

Based on the information provided, what is the mass of the truck?

As a 1250 kg car was passing through an intersection at 8.12 m/s [0.0 degrees] it was hit broadside by a pickup truck running a red light at 8.25 m/s [90.0 degrees]. Immediately after the 0.185 s collision, the velocity of the car was 5.65 m/s [73.1 degrees], and the velocity of the truck was 6.47 m/s [38.5 degrees].

Asked on by angelsings

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justaguide | College Teacher | (Level 2) Distinguished Educator

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An intersection is represented by the x-y axis. A car weighing 1250 kg is moving at 8.12 degrees at an angle equal to 0 degrees. It is hit by a truck moving at 8.25 m/s at an angle equal to 90 degrees. As a result of the collision, the velocity of the car is 5.65 m/s at an angle of 73.1 degrees and the velocity of the truck is 6.47 m/s at an angle of 38.5 degrees.

The collision follows the law of conservation of momentum.

The initial momentum along the x-axis is that of the car is equal to  1250*8.12 = 10150. The initial momentum along the y-axis is that of the truck equal to 8.25*M.

After collision, the momentum along the x-axis is 1250*5.65*cos 73.1 + M*6.47*cos 38.5.

1250*5.65*cos 73.1 + M*6.47*cos 38.5 = 10150

=> M = (10150 - 1250*5.65*cos 73.1)/(6.47*cos 38.5)

=> M = 1600 kg

The momentum along the y-axis is 1250*5.65*sin 73.1 + M*6.47*sin 38.5

1250*5.65*sin 73.1 + M*6.47*sin 38.5 = 8.25*M

=> M = (1250*5.65*sin 73.1)/(8.25 - 6.47*sin 38.5)

=> 1600

The mass of the truck is 1600 kg.

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