Answer:
Force(F) = -80,955.01 N
Explanation:
We need to first determine the impulse that the truck driver received from the car during the collision
So; m₁v₁ - m₂v₂ = (m₁m₂)v₀
where;
m₁ = mass of the truck = 4280 kg
v₁ = v₂ = speed of the each vehicle = 7.69 m/s
m₂ = mass of the car = 810 kg
Substituting our data; we have:
(4280×7.69) - (810×7.69) = (4280+810)v₀
32913.2 - 6228.9 = (5090)v₀
26684.1 = (5090)v₀
v₀ =
v₀ = 5.25 m/s
NOW, Impulse on the truck = m (v₀ - v)
= 4280 × (5.25 - 7.69)
= 4280 × (-2.44)
= -10,443.2 kg. m/s
Force that the seat belt exert on the truck driver can be calculated as:
Impulse = Force × Time
-10,443.2 kg. m/s = F (0.129)
F =
Force(F) = -80,955.01 N
Thus, the Force that the seat belt exert on the truck driver = -80,955.01 N
Answer:
option C
Explanation:
Let mass of the bullet be m and velocity be v
mass of gun be M and bullet be V
now,
using conservation of momentum for gun 1
(M+m) V' = 2 mv + 3 MV
V' = 0
3 M V = - 2 mv
momentum of gun 1 =- 2 mv---------(1)
now for gun 2
(M+m) V' = mv + MV
V' = 0
M V = - mv
momentum of gun 1 = -mv-----------(2)
dividing equation (1) by (2)
the correct answer is option C
Explanation:
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