The city police are in pursuit of Robin Banks after his recent holdup at the savings and loan. The high-speed police chase ends at an intersection as a 2080-kg Ford Explorer (driven by Robin) traveling north at 22.6 m/s collides with a 18400-kg garbage truck moving east at 10.4 m/s. The Explorer and the garbage truck entangle together in the middle of the intersection and move as a single object. Determine the post-collision speed and direction of the two entangled vehicles.

Answers

Answer 1

Answer:

Answer:

$v=9.6215m/s\n\theta=13.8^o$

Explanation:

This problem is an example of a perfectly inelastic collision. After the collision, the two cars merge into a single object. So in order to find the velocity of that object, we need to find the velocity of the center of mass of the given system of 2 cars.

Mass of Ford = $m_F$ = 2080 kg

Mass of truck = $m_T$ = 18400 kg

Velocity of Ford = $v_F$ = 22.6 m/s north

Velocity of truck = $v_T$ = 10.4 m/s east

Let us break up the velocity vector into 2 components, one along north ( $v_N$ ) and one along east ( $v_E$ ).

Therefore,

$v_N=(m_Fv_F_(north)+m_Tv_T_(north))/(m_F+m_T) =((2080*22.6)+(18400*0))/(2080+18400) =2.2953m/s$

(speed of truck along the north is zero)

Similarly,

$v_E=(m_Fv_F_(east)+m_Tv_T_(east))/(m_F+m_T) =((2080*0)+(18400*10.4))/(2080+18400) =9.3437m/s$

(speed of Ford along the east is zero)

Hence, the resultant speed of the entangled cars is given by,

$v=√(v_N^2+v_E^2) =√(2.2953^2+9.3437^2)m/s=9.6215 m/s$

and the direction is given by (please refer to the figure attached),

$\theta=tan^(-1)((v_N)/(v_E) )=tan^(-1)((2.2953)/(9.3437) )=13.8^o$

Answer 2

Answer:

Final answer:

The post-collision speed and direction of the two entangled vehicles, calculated using principles of momentum conservation and trigonometry, are found to be 9.82 m/s and 76.4 degrees east of north respectively.

Explanation:

In this scenario, we are dealing with the principles of momentum conservation. In a collision like this one, the total momentum before the collision equals the total momentum after the collision, as long as no external forces are acting on the system. The momentum (p) of an object is calculated as the mass (m) of the object multiplied by its velocity (v), or p=mv.

Before the collision, the total momentum of the system (ptotal) is the sum of the momentum of the Ford Explorer (pExplorer) and the garbage truck (ptruck). After the collision, they move together as a single object, so their combined mass and velocity would make up the total momentum.

Based on the given information, pExplorer = 2080 kg * 22.6 m/s = 47,008 kg*m/s and ptruck = 18400 kg * 10.4 m/s = 191,360 kg*m/s. The direction should also be taken into account. Since they are perpendicular to each other, we can use the Pythagorean Theorem to find the resultant momentum: (pExplorer)^2 + (ptruck)^2 = ptotal^2, which gives a total momentum of 201,262.8 kg*m/s.

Now, the total mass (mtotal) of two entangled vehicles is 2080 kg + 18400 kg = 20480 kg. Hence the velocity (v) of the two vehicles after the collision is calulated by ptotal divided by mtotal : 201262.8 kg*m/s / 20480 kg = 9.82 m/s. The direction is calculated using trigonometrics such as tan(-1)(ptruck / pExplorer) which gives a direction of 76.4 degrees east of north.

Learn more about Momentum Conservation here:

brainly.com/question/34186969

#SPJ3

Related Questions

suppose that you look into a photometer's eyepiece and the fluorescent disks appear to be equal in intensity. If the distance between the photometer to lamp 1 is 400mm, the distance between the photometer to lamp 2 is 200 mm, and the intensity of lamp 2 is known to be 15 candelas, what is the intensity to lamp 1?
The overall function of the calvin cycle is __________.
A photon with a wavelength of 2.29 × 10^–7 meter strikes a mercury atom in the ground state.Determine the energy, in electronvolts, of this photon.
What is the main force that causes a star to turn into a black hole?Temperature Gravity Pressure Nuclear explosions.
What word is used to describe knowledge about the universe and the method of obtaining that knowledge?

If the breakers at a beach are separated by 5 m and hit shore with a frequency of 0.3 Hz, at what speed are they traveling?

Answers

Wave speed = (frequency) x (wavelength)

= (0.3 Hz) x (5 m)

= 1.5 m-Hz = 1.5 m/s .

A 1600N force is applied to a 85kg mass what is the acceleration of the mass?

Answers

newton 3rd law:
F=ma
1600=85*a
1600/85=a
a=18.82 m/s^2

Heat is a transfer of thermal energy from one object to anther because of a difference in?

Answers

Because of a difference in temperature.

temperature is the answer

You throw a rock out of a 7th story window. you time that it takes 3.7 seconds to hit the ground, and measure that it hit the ground 115 m from the base of the building? How fast must you have thrown the rock?
How high up must the 7th floor be?
If you had thrown it at the same speed but at the 28th floor, how far from the base of the building would it land?

Answers

1) 31.1 m/s

The rock has been thrown straight out of the window: its motion on the horizontal direction is simply a uniform motion, with constant speed $v$ , because no forces act in the horizontal direction. The speed in a uniform motion is given by

$v=(S)/(t)$

where S is the distance traveled and t the time taken.

In this case, the distance by the rock before hitting the ground is $S=115 m$ and the time taken is $t=3.7 s$ , so the initial speed is given by

$v=(115 m)/(3.7 s)=31.1 m/s$

2) 67.1 m

In this part of the problem we are only interested in the vertical motion of the rock. The vertical motion is a uniformly accelerated motion, with constant acceleration $a=9.8 m/s^2$ (acceleration of gravity) towards the ground. In a uniformly accelerated motion, the distance traveled by the object is given by

$S=(1)/(2)at^2$

where t is the time. Substituting a=9.8 m/s^2 and t=3.7 s, we can find S, the vertical distance covered by the rock, which corresponds to the height of the 7th floor:

$S=(1)/(2)(9.8 m/s^2)(3.7 s)^2=67.1 m$

3) 230.1 m

The height of the 7th floor is 67.1 m. So we can assume that the height of each floor is

$h=(67.1 m)/(7)=9.6 m$

And so, the height of the 28th floor is

$h=28\cdot 9.6 m=268.8 m$

We can find the total time of the fall in this case by using the same formula of the previous part:

$S=(1)/(2)at^2$

In this case, S=268.8 m, so we can re-arrange the formula to find t

$t=\sqrt{(2S)/(g)}=\sqrt{(2(268.8 m))/(9.8 m/s^2)}=7.4 s$

And now we can consider the motion of the rock on the horizontal direction: we know that the rock travels at a constant speed of v=31.1 m/s, so the distance traveled is

$S=vt=(31.1 m/s)(7.4 s)=230.1 m$

And this is how far from the building the rock lands.

speed = distance/time taken
115/3.7 =31.08m/s^2

A ball is thrown horizontally from a 28-meter building with a velocity of 5.5 m/s. How far does the ball land from the base of the building? A) 56 m B) 49 m C) 35 m D) 42 m