car a with mass of 1250kg, is traveling at 30 m/s to the east. car b is a truck with mass of 2000kg, traveling to the west at 25m/s. assume these two vehicles experience 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?
yes
To find the resulting velocity of car B after the collision, we need to use the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.
Momentum (p) is calculated by multiplying the mass (m) of an object by its velocity (v): p = m * v.
Before the collision, the momentum of Car A is given by:
pA = mA * vA
= 1250 kg * 30 m/s
= 37500 kg·m/s to the east.
Before the collision, the momentum of Car B is given by:
pB = mB * vB
= 2000 kg * (-25 m/s) [since the velocity is to the west]
= -50000 kg·m/s to the west.
The total momentum before the collision can be calculated by summing up the individual momenta:
pTotal = pA + pB
= 37500 kg·m/s - 50000 kg·m/s
= -12500 kg·m/s to the west.
After the collision, the total momentum is still conserved:
pTotal' = mA' * vA' + mB' * vB'
We are given that car A goes off with a velocity of 10 m/s to the west. So, vA' = -10 m/s.
Substituting the given value into the equation and re-arranging it, we can solve for vB'.
pTotal' = mA' * vA' + mB' * vB'
-12500 kg·m/s = 1250 kg * (-10 m/s) + mB' * vB'
Simplifying the equation further,
-12500 kg·m/s = -12500 kg·m/s + mB' * vB'
To maintain the conservation of momentum, the term on the right-hand side must be 0, so it follows that:
0 = mB' * vB'
Since the mass of car B is 2000 kg, we see that mB' cannot be 0. Therefore, vB' must be 0.
Hence, the resulting velocity of car B after the collision is 0 m/s. Car B will come to a stop.