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andrew-mc [135]
3 years ago
8

In a front-end collision, a 1500 kg car with shock-absorbing bumpers can withstand a maximumforce of 80 000 N before damage occu

rs. If the maximum speed for a non-damaging collision is4.0 km/h, by how much must the bumper be able to move relative to the car
Physics
1 answer:
Harrizon [31]3 years ago
4 0

Answer:

The bumper will be able to move by 0.01155m.

Explanation:

The magnitude of deceleration of the car in the front end collision.

a = \frac{F_m}{m} \\

a = \frac{80000}{1500} \\

a = 53.33

This is the deceleration of the car that is generated to stop due to a front end collision.

4 km/h = 1.11 m/s

Now, the initial speed of the bumper in the relation of car, Vi = 0

Now, the initial speed of the bumper in the relation of car, Vf = 1.11 m/s

Use the below equation:

s = \frac{(Intitial \ speed)^2 – (Final \ speed)^2}{2a} \\

s = \frac{(1.11)^2 – (0)}{2 \times 53.33} \\

s = 0.01155 \\

Thus, the bumper can move relative to the car is 0.01155 m .

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2) The horizontal and vertical components of the initial velocity of a football are 16 m/s and 20 m/s respectively. How long doe
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A 1300 kg car traveling at 35 mph rear-ends a 1000 kg car traveling 25 mph. Just after the collision (but before the driver’s sl
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Answer

given,

before collision

mass of car A = m_a = 1300 kg

velocity of car A = v_a  = 35 mph

mass of car B = m_b= 1000 kg

velocity of car B = v_b  = 25 mph

after collision

V_a = 30 mph

V_b = 31.5 mph

Initial momentum

P_1 = m_av_a + m_b v_b

P_1 = 1300 \times 35+ 1000 \times 25

P_1 =70500 Kg.m/s

final momentum

P_2 = m_aV_a + m_b V_b

P_2 = 1300 \times 30+ 1000 \times 31.5

P_2 =70500 Kg.m/s

here  initial momentum is equal to the final momentum of the car.

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