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3 years ago

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In a physics experiment, two equal-mass carts roll towards each other on a level, low-friction track. One cart rolls rightward a

t 2 m/s and the other cart rolls leftward at 1 m/s. After the carts collide, they couple (attach together) and roll together with a speed of __________. Ignore resistive forces.a. 0.5 m/s
b. 0.33 m/s
c. 0.67 m/s
d. 1.0 m/s
e. none of these

1 answer:

xz_007 [3.2K]3 years ago

7 0

Answer:

The correct answer is option a.

Explanation:

Conservation of momentum :

$m_1u_1+m_2u_2=m_1v_1+m_1v_2$

Where :

$m_1, m_2$ = masses of object collided

$u_1,u_2$ = initial velocity before collision

$v_1,v_2$ = final velocity after collision

We have :

Two equal-mass carts roll towards each other.

$m_1=m_2=M$

Initial velocity of $m_1=u_1=2 m/s$

Initial velocity of $m_2=u_2=-1 m/s$ (opposite direction)

Final velocity of $m_1=v_1=v$ (same direction )

Final velocity of $m_2=v_2=v$ (same direction)

$M\times 2 m/s+M(-1 m/s)=Mv+Mv$

$1 m/s=2v$

v = 0.5 m/s

rg135

The speed of the carts after their collision is 0.5 m/s.

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Answer:

Points downward, and its magnitude is 9.8 m/s^2

Explanation:

The motion of a projectile consists of two independent motions:

- A uniform horizontal motion, with constant velocity and zero acceleration. In fact, there are no forces acting on the projectile along the horizontal direction (if we neglect air resistance), so the acceleration along this direction is zero.

- A vertical motion, with constant acceleration g = 9.8 m/s^2 towards the ground (downward), due to the presence of gravity wich "pulls" the projectile downward.

The total acceleration of the projectile is given by the resultant of the horizontal and vertical components of the acceleration. But we said that the horizontal component is zero, therefore the total acceleration corresponds just to its vertical component, therefore it is a vector with magnitude 9.8 m/s^2 which points downward.

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The length of a simple pendulum is 0.81 mand the mass of the particle (the "bob") at the end of the cable is0.23 kg. The pendulu

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Answer:

$\displaystyle w=3.478\ rad/sec$

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Explanation:

<u>Simple Pendulum</u>

It's a simple device constructed with a mass (bob) tied to the end of an inextensible rope of length L and let swing back and forth at small angles. The movement is referred to as Simple Harmonic Motion (SHM).

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We have the length of the pendulum is L=0.81 meters, then we have

$\displaystyle w=\sqrt{\frac{9.8}{0.81}}$

$\displaystyle w=3.478\ rad/sec$

(b) The total mechanical energy is computed as the sum of the kinetic energy K and the potential energy U. At its highest point, the kinetic energy is zero, so the mechanical energy is pure potential energy, which is computed as

$U=mgh$

where h is measured to the reference level (the lowest point). Please check the figure below, to see the desired height is denoted as Y. We know that

$H+Y=L$

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$H=L\ cos\alpha$

Solving for Y

$Y=L(1-cos\alpha )$

$Since\ \alpha=8.1^o, L=0.81\ m$

$Y=0.0081\ m$

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$U=mgh=0.23\ kg(9.8\ m/s^2)(0.0081\ m)$

$U=0.0182\ J$

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$M=K+U=0+U=U$

$M=0.0182\ J$

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$\displaystyle K=\frac{mv^2}{2}$

Equating to the mechanical energy of the system (M)

$\displaystyle \frac{mv^2}{2}=0.0182$

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$\displaystyle v=\sqrt{\frac{(2)(0.0182)}{0.23}}$

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Lena [83]

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<h3>Angular velocity of the tire</h3>

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The magnitude of the angular velocity of the tire is calculated as follows;

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