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N76 [4]
3 years ago
5

A small object is sliding along a level surface with negligible friction, and has mass m, and constant speed v_0, when it collid

es with a brick resting on the surface. If the object recoils backward at (v_0)/3, and is in contact with the brick for time delta t, what average force did the brick exert on the object? Assume the object is moving along x-direction.
A. F_net = 0
B. F_net= -((mv_0)/delta t)
C. F_net = -((4mv)/(3delta t))
D. F_net = -((mv_0)/ delta t)
E. F_net = -((3mv_0)/(2delta t))
F. F_net = -((2mv_0)/(3delta t))
Physics
1 answer:
VMariaS [17]3 years ago
7 0

To solve this problem we will apply the concepts related to the Impulse-Momentum theorem. On this theorem it is specified that the Momentum is defined as the product between the mass and the velocity of the object, and the Impulse as the product between the Force and time. Both concepts fully comparable to each other. Mathematically they can be described as,

\Delta p = m \Delta v

And

\Delta P = F_{net} \Delta t

Here,

m = mass\\\Delta v = \text{Change in velocity}\\F_{net} = \text{ Net Force}\\\Delta t = \text{Change in time}\\

According to the value given we have that the change in velocity in the momentum is

\Delta p = m(-v_0 - \frac{v_0}{3})

\Delta p = -4 m \frac{(v_0)}{3}

Using this value to find the Net force we have that

F_{net} = \frac{\Delta P}{\Delta t}

F_{net} =\frac{-4 m \frac{v_0}{3}}{\Delta t}

F_{net}  = \frac{-4 mv_0}{3 \Delta t}

Therefore the correct answer is C.

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Inertia depends on mass, the more mass the more inertia.
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A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th
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a) x(t)=2.0 sin (10 t) [m]

The equation which gives the position of a simple harmonic oscillator is:

x(t)= A sin (\omega t)

where

A is the amplitude

\omega=\sqrt{\frac{k}{m}} is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s

The amplitude, A, can be found from the maximum velocity of the spring:

v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m

So, the equation of motion is

x(t)= 2.0 sin (10 t) [m]

b)  t=0.10 s, t=0.52 s

The potential energy is given by:

U(x)=\frac{1}{2}kx^2

While the kinetic energy is given by:

K=\frac{1}{2}mv^2

The velocity as a function of time t is:

v(t)=v_{max} cos(\omega t)

The problem asks as the time t at which U=3K, so we have:

\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}

However, \frac{m}{k}=\frac{1}{\omega^2}, so we have

(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\

with two solutions:

\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s

\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s

c) 3 seconds.

When x=0, the equation of motion is:

0=A sin (\omega t)

so, t=0.

When x=1.00 m, the equation of motion is:

1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s

The period of a pendulum is:

T=2 \pi \sqrt{\frac{L}{g}}

where L is the length of the pendulum. By using T=0.628 s, we find

L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m






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