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zhuklara [117]
3 years ago
10

You have been assigned to investigate a traffic accident. The masses of car A and car B are 1300 kg and 1200 kg, respectively. C

ar B was initially at rest with all four wheels locked. Skid marks indicate car B slid 10 m after the impact. Assuming that the coefficient of kinetic friction the tires and road is 0.8 and the coefficient of restitution of the impact is 0.4, calculate the velocity of A just before impact.
Physics
1 answer:
jarptica [38.1K]3 years ago
4 0

Answer:

The velocity of A before impact = 17.90 m/s

Explanation:

Coefficient of restitution = (speed of seperation)/(speed of approach)

= (v₁ - v₂)/(u₂ - u₁)

where v₁ = velocity of the car A after the impact = ?

v₂ = velocity of the car B after the impact = ?

u₂ = velocity of the car B before the impact = 0 m/s (it was initially at rest)

u₁ = velocity of car A before the impact = ?

First of, we can solve for v₂, the velocity of car B after the impact, from some of the information given in the question.

- Skid marks indicate car B slid 10 m after the impact

- The coefficient of kinetic friction the tires and road is 0.8.

According to the work energy theorem, the work done by frictional force in stopping the car B is equal to the change in kinetic energy of the car B. (All after collision)

W = ΔK.E

ΔK.E = (1/2)(1200)(v₂²) - 0 (final kinetic energy is 0 since the car comes to stop eventually)

ΔK.E = (600v₂²) J

W = F × d

where F = frictional force = μmg = 0.8×1300×9.8 = 10,192 N

d = distance the car skids over before stopping = 10 m

W = 10,192 × 10 = 101,920 J

W = ΔK.E

101,920 = 600v₂²

v₂² = (101920/600) = 169.867

v₂ = 13.03 m/s

But recall,

Coefficient of restitution = (v₁ - v₂)/(u₂ - u₁)

For the sake of convention, we take the direction of car A's initial velocity to be the positive direction.

u₁ = ?

u₂ = 0 m/s

v₁ = ?

v₂ = +13.03 m/s

Coefficient of restitution = 0.4

0.4 = (v₁ - 13.03)/(0 - u₁)

-0.4u₁ = v₁ - 13.03

v₁ = 13.03 - 0.4u₁

But this is a collision. In a collision, the linear momentum is usually conserved.

Momentum before collision = Momentum after collision

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

1300u₁ + (1200×0) = 1300v₁ + (1200×13.03)

1300u₁ + 0 = 1300v₁ + 15639.95

1300u₁ = 1300v₁ + 15639.95

But recall, from the coefficient of restitution relation,

v₁ = 13.03 - 0.4u₁

Substituting this into the momentum balance equation.

1300u₁ = 1300v₁ + 15639.95

1300u₁ = 1300(13.03 - 0.4u₁) + 15639.95

1300u₁ = 16943.28 - 520u₁ + 15639.95

1820u₁ = 32,583.23

u₁ = (32,583.23/1820)

u₁ = 17.90 m/s

Therefore, the velocity of A before impact = 17.90 m/s

Hope this Helps!!!

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FrozenT [24]
I would think the answer is c.
5 0
3 years ago
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A 28 kg mass suspends from a light rope 18 m long & is held to one side by the horizontal force, F, as shown below.
frutty [35]

Answer: 215.15 N

Explanation:

If we draw a free body diagram of the mass we will have the following:

\sum{F_{x}}=-Tcos\theta + F=0 (1)

\sum{F_{y}}=Tsin\theta - mg=0 (2)

Where T is the tension force of the rope, m=28 kg the mass, g=9.8 m/s^{2} the acceleration due gravity and mg is the weight.

On the other hand, we can calculate \theta as follows:

cos\theta=\frac{s}{l}

\theta=cos^{-1}(\frac{s}{l})

Where s=11.1 m and l=18 m

\theta=cos^{-1}(\frac{11.1 m}{18 m})

\theta=51.9\° (3)

Now, we firstly need to find T from (2):

T=\frac{mg}{sin\theta} (4)

T=\frac{(28 kg)(9.8 m/s^{2})}{sin(51.9\°)}

T=348.69 N (5)

Substituting (5) in (1):

F=Tcos\theta (6)

F=348.69 N cos(51.9\°)

Finally:

F=215.15 N

8 0
3 years ago
. On a safari, a team of naturalists sets out toward a research station located 9.6 km away in a direction 42° north of east. Af
Elden [556K]

Answer:\theta =49.76^{\circ} North of east

Explanation:

Given

Research station is 9.6 km away in 42^{\circ}North of east

after travelling 3.1 km 25^{\circ} north of east

Position vector of safari after 3.1 km is

r_2=3.1cos25\hat{i}+3.1sin25\hat{j}

Position vector if had traveled correctly is

r_0=9.6cos42\hat{i}+9.6sin42\hat{j}

Now applying triangle law  of vector addition we can get the required vector(r_1)

r_1+r_2=r_0

r_1=(9.6cos42-3.1cos25)\hat{i}+(9.6sin42-3.1sin25)\hat{j}

r_1=4.325\hat{i}+5.112\hat{j}

Direction is given by

tan\theta =\frac{y}{x}=\frac{5.112}{4.325}

\theta =49.76^{\circ}

8 0
3 years ago
A vehicle of mass 1600kg take off from rest and coveres a distance of 280m in 60s. It continues at this speed for 98seconds and
Lady bird [3.3K]
1. V = Vo + a*t. 
<span>15 = 0 + a*5 a = 3 m/s^2. </span>

<span>2. V = Vo + a*t. </span>
<span>0 = 15 - 5*t, t = 3 s. </span>

<span>3. d1 = Vo*t + 0.5a*t^2. </span>
<span>Vo = 0, t = 5, a = 3 m/s^2, d1 = ?. </span>

<span>d2 = Vo*t. Vo = 15 m/s, t = 15 s, </span>
<span>d2 = ?. </span>

<span>V^2 = Vo^2 + 2a*d3. </span>
<span>V = 0, Vo = 15 m/s, a = -5 m/s^2, d3 = ?. </span>

<span>d1+d2+d3 = Total dist.</span>
4 0
3 years ago
Alternate current
dusya [7]

sorry bro but i don't know his i am also bad

Explanation:

but maybe you can you the formula of volatage as well as

5 0
2 years ago
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