Physics, 11.03.2020 00:59 bryson3307

A family car has a mass of 1400 kg. In an accident it hits a wall and goes from a speed of 27 m/s to a standstill in 1.5 seconds. By how much would the force have been reduced if the car had had a crumple zone that increased the collision time to 2.2 seconds? Give your answer to the nearest whole number.

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Answer from: alliemeade1

The amount of force which is reduced is required.

The force would be reduced by 8018.18 N.

m = Mass of car = 1400 kg

u = Initial velocity = 27 m/s

v = Final velocity = 0 m/s

$\Delta t_1$ = Initial time = 1.5 s

$\Delta t_2$ = Final time = 2.2 s

Force is given by

$F=m\dfrac{v-u}{\Delta t}$

The difference in force will be

$\Delta F=\left|m(v-u)\left(\dfrac{1}{\Delta t_1}-\dfrac{1}{\Delta t_2}\right)\right|\\\Rightarrow \Delta F=\left|1400(0-27)\left(\dfrac{1}{1.5}-\dfrac{1}{2.2}\right)\right|\\\Rightarrow \Delta F=8018.18\ \text{N}$

The force would be reduced by 8018.18 N.

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Answer from: themajesty9898

The force has been reduced by 8018 N

Explanation:

The impulse exerted on the car during the crash is equal to the product of the force exerted and the duration of the collision, and it is also equal to the change in momentum of the car. So we can write:

$F\Delta t = m\Delta v$

where:

F is the force exerted on the car

$\Delta t$ is the duration of the collision

m = 1400 kg is the mass of the car

$\Delta v=-27 m/s$ is the change in velocity of the car

We can re-write the equation as

$F=\frac{m\Delta v}{\Delta t}$

In the 1st collision, the time is 1.5 seconds, so the force is

$F_1=\frac{(1400)(-27)}{1.5}=-25,200 N$

In the 2nd collision, the time is increased to 2.2 seconds, so the force is

$F_2=\frac{(1400)(-27)}{2.2}=-17,182 N$

Therefore, the force has been reduced by:

F_2-F_1=-17,182-(-25,200)=8018 N

Answer from: Quest

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Answer from: Quest

m₁v₁ + m₂v₂ = (m₁ + m₂)vmomentum is conserved.the left-hand side of the equation represents the momentum in the system before the collision.the right-hand side represents the momentum in the system after the collisionm₁ and v₁ represents the mass and the velocity of the physics blockm₂ and v₂ represents the mass and velocity of the ball of puttyv represents the final velocity of the combined mass.the question states the two masses stick together. they combined mass has its own velocity with the body mass the sum of the two objects' massestheir combined momentum is (m₁ + m₂)vto find the magitude of (m₁ + m₂)v, we need to add the vectors on the left-hand side of the equation: m₁v₁ + m₂v₂.note the diagram. i have positioned the vectors in accordance with their directions.the magnitude of the resultant can be calculated using the pythagorean theorem since this is a right triangle situation (north and west are perpendicular)|(m₁ + m₂)v| = √[ (m₁v₁)² + (m₂v₂)² ]|(m₁ + m₂)v| = √[ (1.5kg·8.0m/s)² + (0.40kg·20m/s)² ]|(m₁ + m₂)v| = 14.4 kg m/sfor two significant figures (assuming that 20m/s is two), this rounds to 14 kg m/s

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A family car has a mass of 1400 kg. In an accident it hits a wall and goes from a speed of 27 m/s to...

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