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# Tutor profile: Mujibur B.

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Mujibur B.
I have been tutoring full time for 5 years. Now I am doing it as a part-time as a second job
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## Questions

### Subject:Physics (Newtonian Mechanics)

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Question:

What is the coefficient for friction A 6-g bullet is shot into a 2.1-kg block of wood, where it becomes embedded. The block is initially at rest and the bullet has an initial velocity of 400 m/s. The block then travels 1.2 m horizontally. What is the coefficient of kinetic friction between the block and the surface?

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Mujibur B.

Let's first build the solution logically in reverse! Steps: To find the coefficient of kinetic friction using the formula $$F= \mu N$$ , we have to find how much frictional force was applied to stop the block with the bullet. To find the force F, we have to find the work done by the force. WHich is equal to the change in Kinetic Energy of the bullet. The bullet embedded in the block came to a complete stop. So, the change in kinetic energy is equal to the initial kinetic energy. We need to get the initial velocity of the block with the bullet. To get the velocity of the block with the bullet, we can use the conservation of momentum with the bullet and block collision. Now let's start from the last step and reach our goal of finding the coefficient of kinetic friction First, we have to consider the momentum conservation mass of bullet, $$m_1 = 6g =0.006 kg$$ initial velocity of the bullet,$$v_1 =400 ms^{-1}$$ mass of block, $$m_2 = 2.1 kg$$ initial velocity of the block, $$v_2=0 ms^{-1}$$ final mass, $$m_3= 2.106 kg$$ what is the final velocity of the block with bullet, $$v_3$$? Using Momentum conservation, $$m_1 v_1 + m_2 v_2 = m_3 v_3 \\ \Rightarrow 0.006kg \times 400ms^{-1} + 2.1 \times 0kg ms^{-1} = 2.106kg \times v_3\\ \Rightarrow 2.4 kg ms^{-1} =2.106kg \times v_3\\ \Rightarrow v_3 = 1.1396 ms^{-1}$$ We can calculate the kinetic energy of the block with bullet from this Kinetic Energy is, $$E = 1/2 m v^2 \\= 0.5 \times m_3 \times (v_3)^2\\ = 1.3675 kg m^2s^{-2}\\ = 1.3675 J$$ Let's say that the Frictional force that stopped the bullet is F, then Work done by this force= Kinetic Energy of the Block with Bullet Since the force is a frictional force, we can say it was a constant force and work done by it is work, $$W = F x$$ , where x is the distance traveled by the Block with the bullet. So, $$x= 1.2m$$ Now, $$Fx = W = Kinetic Energy = 1.3675 J$$ so, $$F \times 1.2m =1.3675 J$$ so, $$F =\frac{ 1.3675 J }{ 1.2 m}\\ = 1.1396 Kg m s^{-2}\\ = 1.1396 N$$ We know the frictional force, $$F = \mu N$$ where $$\mu$$ is the coefficient of friction and N is the Normal Force Since the surface is horizontal, the weight of the block with bullet is equal to the magnitude of the normal force. $$N = m_3 \times$$ gravitational acceleration So, $$F =\mu \times 2.106 \times 9.8 N$$ so, $$\mu = \frac{1.1396} {20.6388}\\ = 0.055$$ Coefficient of kinetic friction is 0.055

### Subject:Physics (Electricity and Magnetism)

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Question:

Particle colliders The question is as follows In the Large Electron-Positron collider at CERN, groups of electron and positrons were accelerated along a circular tunnel, so that they collided. Electromagnets were used to keep the particles moving in a circle. Discuss how the strength and direction of the magnetic field would need to be adjusted to i) keep a particle travelling in a circle while it is increasing its speed ii) bend the paths of both positrons and electrons which are travelling in opposite directions in the same tunnel.

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Mujibur B.

The centripetal force required to keep the particle in a circular orbit is $$\vec{F}_C = m \frac{v^2}{r}\hat{r}$$ ⋅ ⋅⋅⋅⋅⋅⋅⋅ ⋅⋅ ⋅⋅ ⋅ ⋅⋅⋅⋅⋅ ⋅⋅⋅⋅ ⋅⋅⋅⋅ eq 1 here, m = mass of the particle v = velocity of the particle r = radius of the orbit Lorentz force applied by the magnetic field is $$\vec F_L = q\vec v \times \vec B$$ ⋅ ⋅⋅⋅ ⋅⋅⋅ ⋅⋅⋅ ⋅ ⋅⋅⋅⋅⋅⋅ ⋅⋅⋅⋅ ⋅⋅⋅⋅ ⋅⋅ eq 2 Here to keep the forces in equilibrium, $$\vec F_L= \vec F_C$$ If we apply the right hand rule, we will see that magnetic field perpendicular to the plane of orbit creates a Lorentz force directed always to the center. That makes the angle between v and B $$90^o.$$ So, $$|| \vec v \times \vec B || = vB$$ now if we balance the magnitude of the forces, $$qvB = m\frac{v^2}{r} \\ \Rightarrow B = m\frac{v}{ r q}$$ ⋅ ⋅⋅⋅⋅ ⋅ ⋅⋅⋅⋅⋅ ⋅⋅⋅⋅⋅ ⋅⋅⋅ eq 3 Now to answer the 1st question i) keep a particle traveling in a circle while it is increasing its speed Assuming that we will change only velocity and magnetic field, we differentiate both sides with respect to time, $$\frac{d}{dt} B = \frac{m}{ r q } × (\frac{d}{dt} v) \textit{ ---------------- eq. 4}\\ \Rightarrow \frac{d}{dt} B = constant × (\frac{d}{dt} v) \\ \Rightarrow \frac{d}{dt} B ∝ \frac{d}{dt} v$$ so, you will have to increase the magnetic field in the same rate with the velocity ii) bend the paths of both positrons and electrons which are travelling in opposite directions in the same tunnel. If we look at eq-4 , we can see that if we replace q with -q and v with -v , there is no change in magnetic field or anything else. That means, for a particle with same mass but with opposite charge going at the opposite direction will also stay in the orbit.

### Subject:Physics

TutorMe
Question:

What is the coefficient for friction A 6-g bullet is shot into a 2.1-kg block of wood, where it becomes embedded. The block is initially at rest and the bullet has an initial velocity of 400 m/s. The block then travels 1.2 m horizontally. What is the coefficient of kinetic friction between the block and the surface?

Inactive
Mujibur B.

Let's first build the solution logically in reverse! Steps: To find the coefficient of kinetic friction using the formula $$F= \mu N$$ , we have to find how much frictional force was applied to stop the block with the bullet. To find the force F, we have to find the work done by the force. WHich is equal to the change in Kinetic Energy of the bullet. The bullet embedded in the block came to a complete stop. So, the change in kinetic energy is equal to the initial kinetic energy. We need to get the initial velocity of the block with the bullet. To get the velocity of the block with the bullet, we can use the conservation of momentum with the bullet and block collision. Now let's start from the last step and reach our goal of finding the coefficient of kinetic friction First, we have to consider the momentum conservation mass of bullet, $$m_1 = 6g =0.006 kg$$ initial velocity of the bullet,$$v_1 =400 ms^{-1}$$ mass of block, $$m_2 = 2.1 kg$$ initial velocity of the block, $$v_2=0 ms^{-1}$$ final mass, $$m_3= 2.106 kg$$ what is the final velocity of the block with bullet, $$v_3$$? Using Momentum conservation, $$m_1 v_1 + m_2 v_2 = m_3 v_3 \\ \Rightarrow 0.006kg \times 400ms^{-1} + 2.1 \times 0kg ms^{-1} = 2.106kg \times v_3\\ \Rightarrow 2.4 kg ms^{-1} =2.106kg \times v_3\\ \Rightarrow v_3 = 1.1396 ms^{-1}$$ We can calculate the kinetic energy of the block with bullet from this Kinetic Energy is, $$E = 1/2 m v^2 \\= 0.5 \times m_3 \times (v_3)^2\\ = 1.3675 kg m^2s^{-2}\\ = 1.3675 J$$ Let's say that the Frictional force that stopped the bullet is F, then Work done by this force= Kinetic Energy of the Block with Bullet Since the force is a frictional force, we can say it was a constant force and work done by it is work, $$W = F x$$ , where x is the distance traveled by the Block with the bullet. So, $$x= 1.2m$$ Now, $$Fx = W = Kinetic Energy = 1.3675 J$$ so, $$F \times 1.2m =1.3675 J$$ so, $$F =\frac{ 1.3675 J }{ 1.2 m}\\ = 1.1396 Kg m s^{-2}\\ = 1.1396 N$$ We know the frictional force, $$F = \mu N$$ where $$\mu$$ is the coefficient of friction and N is the Normal Force Since the surface is horizontal, the weight of the block with bullet is equal to the magnitude of the normal force. $$N = m_3 \times$$ gravitational acceleration So, $$F =\mu \times 2.106 \times 9.8 N$$ so, $$\mu = \frac{1.1396} {20.6388}\\ = 0.055$$ Coefficient of kinetic friction is 0.055

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