A 1459 kg car is traveling WEST at 43 m/s. A 9755 kg truck is traveling EAST at 11 m/s. They collide head-on, and stick together. Assuming EAST to be the positive direction, what is the velocity after this collision?

answer in m/s

Both vehicles move east at 3.97 m/s

Explanation:

Law Of Conservation Of Linear Momentum

It states that the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is:

P=mv.

If we have a system of two bodies, then the total momentum is the sum of both momentums:

If a collision occurs and the velocities change to v', the final momentum is:

Since the total momentum is conserved, then:

P = P'

Assume both masses stick together after the collision at a common speed v', then:

The common velocity after this situation is:

Assuming east direction to be positive, we have an m1=1459 kg car traveling west at v1=-43 m/s. An m2=9755 kg truck is traveling east at v2=11 m/s. They collide head-on and stick together after that.

Computing the resultant velocity after the collision:

v' = 3.97 m/s

Both vehicles move east at 3.97 m/s

im not to sure what it is

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.