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Maurinko [17]
3 years ago
8

A 4.0 kg circular disk slides in the x- direction on a frictionless horizontal surface with a speed of 5.0 m/s as shown in the a

djacent Figure. It collides with an identical disk that is at rest before the collision. The collision is elastic. Disk 1 goes off at an 60 5.0 m/s angle of 60 with respect to the x-direction. Disk 2 g 30 goes off at an angle of 30 with respect to the x-direction. What best describes the speeds of the disks after the collision?
Physics
1 answer:
finlep [7]3 years ago
5 0

Solution :

Let $m_1=m_2=4$ kg

$u_1 = 5$ m/s

Let $v_1$ and $v_2$ are the speeds of the disk $m_1$ and $m_2$  after the collision.

So applying conservation of momentum in the y-direction,

$0=m_1 .v_1_y -m_2 .v_2_y $

$v_1_y = v_2_y$

$v_1 . \sin 60=v_2. \sin 30$

$v_2 = v_1 \times \frac{\sin 60}{\sin 30}$

$v_2=1.732 \times v_1$

Therefore, the disk 2 have greater velocity and hence more kinetic energy after the collision.

Now applying conservation of momentum in the x-direction,

$m_1.u_1=m_1.v_1_x+m_2.v_2_x$

$u_1=v_1_x+v_2_x$

$5=v_1. \cos 60 + v_2 . \cos 30$

$5=v_1. \cos 60 + 1.732 \times v_1 \cos 30$

$v_1 = 2.50$ m/s

So, $v_2 = 1.732 \times 2.5$

          = 4.33 m/s

Therefore, speed of the disk 2 after collision is 4.33 m/s

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we use this formula
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3 years ago
Attached here is the question:
ICE Princess25 [194]

Answer:

t = 6 [s]

Explanation:

In order to solve this problem we must first use this equation of kinematics.

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First we must convert the velocity from kilometers per hour to meters per second.

72 [\frac{km}{h}]*\frac{1000m}{1km} *\frac{1h}{3600s} =20 [m/s]

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Now using this other equation of kinematics.

v_{f}=v_{o}-a*t

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43.16°

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The relation between the distance of the slits and the angle through which it is passed is:

dsinθ=nλ

As d and λ are constant

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