Question

A 130 kg truck with a velocity of 12m / s crashes into a parked car with a mass of 100 kgWhat is the final velocity of the two objects if they stick together after the collision

Answer 1

Step-by-step explanation:

m1 of the truck: 130 kg

m2 of the car: 100 kg

v1 of the truck: 12 m/s

vf?

use the formula:

(m1×v1) + (m2×v2) = (m1+m2)vf

so

`vf = ((m1 * v1) + (m2 * v2))/((m1 + m2))`

`vf = ((130 * 12) + (100 * 0))/((130 + 100))`

vf = 1560 ÷ 230 = 6,7 m/s

Answer 2

6.78 m/s

Step-by-step explanation:

We can use the law of conservation of momentum to solve this problem. According to this law, the total momentum of a system is conserved in the absence of external forces. In this case, the system consists of the truck and the parked car.

Before the collision, the momentum of the truck is given by:

p1 = m1 * v1 = 130 kg * 12 m/s = 1560 kg*m/s

The momentum of the parked car is zero since it is not moving.

The total momentum of the system before the collision is therefore:

p1_total = p1 + p2 = 1560 kg*m/s

After the collision, the two objects stick together and move with a common velocity, which we will call v_final. The total mass of the system is:

m_total = m1 + m2 = 130 kg + 100 kg = 230 kg

The total momentum of the system after the collision is:

p2_total = m_total * v_final

According to the law of conservation of momentum, p1_total = p2_total. Therefore:

1560 kg*m/s = 230 kg * v_final

Solving for v_final, we get:

v_final = 1560 kg*m/s / 230 kg = 6.78 m/s

Therefore, the final velocity of the two objects after the collision is 6.78 m/s.