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3 years ago

14

When a 5.0-kilogram cart moving with a speed of 2.8 meters per second on a horizontal surface collides with a 2.0 kilogram cart

initially at rest, the carts locked together. What is the speed of the combined carts after the collision

1 answer:

mixer [17]3 years ago

8 0

Here in all such collision type question we can use momentum conservation as we can see that there is no external force on this system

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

as we know that

$m_1 = 5 kg$

$m_2 = 2 kg$

$v_{1i} = 2.8 m/s$

$v_{2i} = 0 m/s$

now from above equation we have

$5(2.8) + 2(0) = m_1v+ m_2v$

$14 = (5+ 2) v$

$v = 2 m/s$

so the speed of combined system is 2 m/s

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3 years ago

An object is originally moving at a constant velocity of 8 m/s in the -x direction. It moves at this constant velocity for 3 sec

aivan3 [116]

Answer:

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After this, the ball will start traveling with a constant acceleration motion. Due to the fact that the acceleration is the opposite direction to the initial velocity, this motion will have 2 phases:

1. The velocity will start to decrease untill it reaches 0m/s.

2. Then, the velocity will start to increase at the rate of the acceleration.

The distance that the ball travels in the first phase can be found with the following expression:

$v^2 = v_0^2 + 2a*d$

Where v is the final velocity (0m/s), v_0 is the initial velocity (-8m/s) and a is the acceleration (+9m/s^2). We solve for d:

$d = \frac{v^2 - v_0^2}{2a} = \frac{(0m/s)^2 - (-8m/s)^2}{2*7m/s^2}= -4.57m$

Now, before finding the distance traveled in the second phase, we need to find the time that took for the velocity to reach 0:

$t_1 = \frac{v}{a} = \frac{8m/s}{7m/s^2} = 1.143 s$

Then, the time of the second phase will be:

$t_2 = 9s - t_1 = 9s - 1.143s = 7.857s$

Using this, we using the equations for constant acceleration motion in order to calculate the distance traveled in the second phase:

$x = \frac{1}{2}a*t^2 + v_0*t + x_0$

V_0, the initial velocity of the second phase, will be 0 as previously mentioned. X_0, the initial position, will be 0, for simplicity:

$x = \frac{1}{2}*7\frac{m}{s^2}*t^2 + 0m/s*t + 0m = 216.07m$

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2 years ago

A bowling ball is dropped off the top of the Eiffel Tower. If the Eiffel Tower is 300 meters

patriot [66]

Answer:

7.82 s

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Given:

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3 years ago

A neutral solid metal sphere of radius 0.1 m is at the origin, polarized by a point charge of 2 × 10−8 C at location m. At locat

liraira [26]

Answer: E = $1.8 *10 ^{4} N$

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2 years ago

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