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lidiya [134]
3 years ago
11

Two identical sticky masses m are moving in the xy-plane, with their momenta at an angle of φ with one another. They are each mo

ving at the same speed v when they collide at the origin of the coordinates and stick together. After the collision, the masses move at an angle −θ2 with respect to the +x axis at speed v2 .1. What was the angle φ?
Physics
1 answer:
postnew [5]3 years ago
8 0

Answer:

ucosφ=-v2cosθ2\\\\φ=cos^{-1} (\frac{-v2cosθ2}{cosφ} )

Explanation:

Two identical sticky masses m are moving in the xy-plane, with their momenta at an angle of φ with one another. They are each moving at the same speed v when they collide at the origin of the coordinates and stick together. After the collision, the masses move at an angle −θ2 with respect to the +x axis at speed v2 .1. What was the angle φ?

from the principle of momentum

In a system of colliding bodies,we know that the total momentum before collision will equal to the total momentum after collision.

Take note that momentum is the product of mass and velocity

momentum before collision=momentum after collision

mass, m

u=initial velocity of the identical masses

v2=the common velocity after the collision

Note that the collision is inelastic , since they both moved with the same velocity

umcosφ+umcosφ=(m+m)v2cos−θ2

2mucosφ=2mv2cos−θ2

ucosφ=-v2cosθ2\\\\φ=cos^{-1} (\frac{-v2cosθ2}{cosφ} )

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(kindly find attachment below)

From the attachment below, it can be seen that the resultant displacement and the other 2 displacements form a right angle triangle, with A+B as the hypotenus, 3.2km as the opposite and the displacement B as the adjacent.

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3 years ago
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Using formula of the torque act on the wheel

\tau=I\alpha

\alpha=\dfrac{\tau}{I}

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\alpha=\dfrac{3.00}{5.00}

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Using formula of angular velocity

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\omega_{f}=\omega_{i}+\alpha t

Put the value into the formula

\omega_{f}=0+0.6\times8.00

\omega_{f}=4.8\ rad/s

We need to calculate the rotational kinetic energy of the wheel

Using formula of the rotational kinetic energy

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K.E_{rot}=\dfrac{1}{2}\times5.00\times(4.8)^2

K.E_{rot}=57.6\ J

Hence, The wheel's rotational kinetic energy is 57.6 J.

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