- The final velocity of the objects is 229.82km/hr
- Since energy was lost after the collision, the type of collision that occurs is an elastic collision.
- The required force is 1,090,918,000N
- The amount of Kinetic energy lost is 12,401,976.656Joules
According to the law of collision'
m1u1 + m2u2 = (m1+m2)v
Given the following:
m1 = 22,680kg
u1 =170km/h
m2 = 1200kg
u2 = 5km/hr
Get the final velocity "v"
22680(170) + 1200(5) = (22680 + 1200)v
3855600 + 6000 = 12880v
3,861,600 = 12880v
v = 3,861,600/12,880
v = 229.82km/hr
Hence the final velocity of the objects is 229.82km/hr
Since energy was lost after the collision, the type of collision that occurs is an elastic collision.
According to Newton's second law, the formula for calculating the force is expressed as;
F = ma
F = m(v-u)/t
F = 22680+1200(229.82-175)/t
Ft = 23880(54.82)
F = 1,309,101.6/0.0012
F = 1,090,918,000N
Hence the required force is 1,090,918,000N
KE lost = Kinetic energy after collision - Kinetic energy before collision
Kinetic energy after collision = 1/2 * 12880 * 229.82²
Kinetic energy after collision = 340,142,976.656 Jolues
Kinetic energy before collision = 1/2 * 22680(170)²+ 1/2*1200(5)²
Kinetic energy before collision = 327,726,000 + 15000
Kinetic energy before collision = 327,741,000
Kinetic energy lost = 340,142,976.656 - 327,741,000 = 12,401,976.656Joules
Learn more here: brainly.com/question/9537044
