Car a with a mass of 1250kg is travelling at 30m/s to the east. Car b is a truck with a mass of 2000kg, travelling to the west at 25m/s. Assume these two vehicles experienc an inelastic collision but do not stick together and car a goes off 10m/s to the west what will be the resulting velocity of car b
To find the resulting velocity of Car B after the collision, we can use the principle of conservation of momentum. The total momentum before the collision should be equal to the total momentum after the collision.
Momentum (p) is defined as the product of mass (m) and velocity (v): p = m * v.
Before the collision, the total momentum of the system is given by:
Total momentum before = momentum of Car A + momentum of Car B
momentum of Car A = mass of Car A * velocity of Car A
= 1250 kg * 30 m/s (to the east)
momentum of Car B = mass of Car B * velocity of Car B
= 2000 kg * (-25 m/s) (to the west)
Total momentum before = (1250 kg * 30 m/s) + (2000 kg * (-25 m/s))
= 37500 kg*m/s - 50000 kg*m/s
= -12500 kg*m/s
After the collision, the total momentum should still be the same, but the sign will change as Car A is now going to the west:
Total momentum after = momentum of Car A + momentum of Car B
momentum of Car A = mass of Car A * velocity of Car A (after the collision)
= 1250 kg * (-10 m/s) (to the west)
momentum of Car B = mass of Car B * velocity of Car B (after the collision)
= 2000 kg * v (to be determined)
Total momentum after = (1250 kg * (-10 m/s)) + (2000 kg * v)
Since momentum is conserved, we can equate the total momentum before and after the collision:
-12500 kg*m/s = (1250 kg * -10 m/s) + (2000 kg * v)
Now we can solve for the velocity of Car B (v):
-12500 kg*m/s = -12500 kg*m/s + (2000 kg * v)
By canceling out the common terms on both sides of the equation, we can simplify it to:
0 = 2000 kg * v
Since there is no term containing v remaining, the velocity of Car B after the collision is 0 m/s. This means Car B comes to a complete stop.