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2 years ago

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For a movie stunt, an empty truck with a mass of 2000 kg goes 10 m/s and runs into a stopped car of mass 1000 kg. the truck then

keeps moving and pushes the car along with it. if there are no other forces acting on this system, describe the results of the collision?

2 answers:

sergij07 [2.7K]2 years ago

7 0

Explanation:

Given that,

Mass of the truck, m₁ = 2000 kg

Mass of the stopped car, m₂ = 1000 kg

Initial speed of the truck, u₁ = 10 m/s

Initial speed of the car, u₂ = 0 (at rest)

As there is no other forces acting on this system, the momentum remains constant. Let v is the combined speed of the system after the collision. So,

$m_1u_1+m_2u_2=(m_1+m_2)v$

$2000\times 10+1000\times 0=(1000+2000)v$

v = 6.667 m/s

Therefore, the speed of the combined mass is 6.667 m/s and the speed of the combined mass is less than initial speed.

babunello [35]2 years ago

3 0

ANSWER

Both trucks will move together with speed v = 6.67 m/s

so correct answer will be

The speed of the combined vehicles is less than the initial speed of the truck.

EXPLANATION

As we know that there is no external force on the system of two trucks

So here momentum of the two trucks before collision and after collision will remain same

So here we will have

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

so here we will have

$v_{1i} = 10 m/s$

$v_{2i} = 0$

$m_1 = 2000 kg$

$m_2 = 1000 kg$

now we will have

$2000\times 10 + 1000\times 0 = (2000 + 1000) v$

$v = \frac{20000}{3000} = 6.67 m/s$

so correct answer will be

The speed of the combined vehicles is less than the initial speed of the truck.

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3 years ago

A block (mass = 61.2 kg) is hanging from a massless cord that is wrapped around a pulley (moment of inertia = 1/2MR2 kg · m2, wh

kolezko [41]

Answer:

The angular velocity is $w = 53.35 \ rounds /minute$

Explanation:

From the question we are told that

The mass of the block is $m = 61.2kg$

The of the pulley is $M = 14.2 kg$

The radius of the pulley is $R = 1.5m$

The radius of the cord around the pulley is $r = 1.5 m$

The distance of the block to the floor is $d = 8.0 m$

From the question we are told that the moment of inertia of the pulley is

$I = \frac{1}{2} MR^2 kg \cdot m^2$

Substituting value

$I = \frac{1}{2} * 14.2 * (1.5)^2$

$I = 15.975 kg \cdot m^2$

Using the Newtons law we can express the force acting on the vertical axis as

$ma = mg -T$

=> $T = mg -ma$

Now when the pulley is rotated that torque generated on the massless cord as a r result of the tension T and the radius of the cord around the pulley is mathematically represented as

$\tau = I \alpha$

Here $\alpha$ is the angular acceleration

Here $\tau$ is the torque which can be equivalent to

$\tau = T r$

Substituting this above

$Tr = I \alpha$

Substituting for T

$(mg - ma ) r = I\ r \alpha$

Here $a$ is the linear acceleration which is mathematically represented as

$a = r\alpha$

$(mg - m(r\alpha ) ) r = I\ r \alpha$

$mgr = I\alpha + m(r\alpha ) r$

$mgr = \alpha [ I + mr^2]$

making $\alpha$ the subject

$\alpha = \frac{mgr}{I -mr ^2}$

Substituting values

$\alpha = \frac{61.2 * 1.5 * 9.8}{15.975 + (61.2 ) * (1.5)^2}$

$\alpha =5.854 rad /s^2$

Now substituting into the equation above to obtain the acceleration

$a = 5.854 * 1.5$

$a=8.78 m/s^2$

This acceleration is $a = \frac{v}{t}$

and v is the linear velocity with is mathematically represented as

$v = \frac{d}{t}$

Substituting this into the formula acceleration

$a = \frac{d}{t^2}$

making t the subject

$t = \sqrt{\frac{d}{a} }$

substituting value

$t = \sqrt{\frac{8}{8.78}}$

$t = 0.9545 \ s$

Now the linear velocity is

$v = \frac{8}{0.9545}$

$v = 8.38 m/s$

The angular velocity is

$w = \frac{v}{r}$

So

$w = \frac{8.38}{1.5}$

$w = 5.59 rad/s$

Generally 1 radian is equal to 0.159155 rounds or turns

So 5.59 radian is equal to x

Now x is mathematically obtained as

$x = \frac{5.59 * 0.159155}{1}$

$= 0.8892 \ rounds$

Also

60 second = 1 minute

So 1 second = z

Now z is mathematically obtained as

$z = \frac{ 1}{60}$

z $= 0.01667 \ minute$

Therefore

$w = \frac{0.8892}{0.01667}$

$w = 53.35 \ rounds /minute$

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So, the angle between two vectors having equal magnitude is equal to 120º.

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Answer:

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