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Lerok [7]
2 years ago
5

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.
Physics
1 answer:
horrorfan [7]2 years ago
4 0

Answer:

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

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Answer:

1) 2197.44 J

2) 0 J

3) 2197.44 J = Constant

4) 2197.44 J

5) Approximately 8.86 m/s

Explanation:

The given parameters are;

The mass of the diver, m = 56 kg

The height of the cliff, h = 4.0 m

The speed with which the diver is moving, vₓ = 8.0 m/s

The gravitational potential energy = Mass, m × Height of the cliff, h × Acceleration due to gravity, g

1) Her gravitational potential energy = 56 × 4.0 × 9.81 = 2197.44 J

2) The kinetic energy = 1/2·m·u²

Where;

u = Her initial velocity = 0 when she just leaves the cliff

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Her kinetic energy when she just leaves the cliff = 1/2 × 56 × 0² = 0 J

3) The total mechanical energy = Kinetic energy + Potential energy

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5) The speed with which she enters the water, v, is given from, v² = u² + 2·g·h

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u = The initial velocity at the top of the cliff before she jumps= 0 m/s

∴ v² = 0² + 2 × 9.81 × 4 = 78.48

v = √78.48 ≈ 8.86 m/s

The speed with which she enters the water, v ≈ 8.86 m/s

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