# Tutor profile: Nikhil K.

## Questions

### Subject: Physics (Newtonian Mechanics)

Find the acceleration of a 5kg block attached vertically by string to a 4kg block kept on a frictionless table. Two blocks are connected by a massless string over a massless pulley.

Acceleration "a" can be found by applying Newton's second law along both vertical and horizontal axes and using the fact that tension is the same along both portions of a string. equations can be written as follows (4kg mass = $$m_1$$ and 5kg mass = $$m_2$$) along x-axis- $$ \sum F_x $$ = $$T - m_1g = m_1a$$ $$ \sum F_x $$ = $$T - 4(9.8) = 4a$$ Rearranging for T $$T = (4a + 4(9.8))$$.......................(1) Along y-axis, $$ \sum F_y $$ = $$m_2g-T = m_2a$$ $$ \sum F_y $$ = $$5(9.8)-T = 5a$$..................(2) Substituting for T in (2) from (1) $$5(9.8)-(4a + 4(9.8)) = 5a$$ $$49-4a - 39.2= 5a$$ $$49 - 39.2= 9a$$ $$9.8= 9a$$ $$a = 1.1 m/s^2$$ Hence the vertically attached box of 5kg will accelerate downwards with acceleration $$1.1m/s^2$$ **Next step to think about** Adding the friction to the horizontal surface will alter $$ \sum F_x $$ equation to $$ \sum F_x $$ = $$T - m_1g-F_\mu = m_1a$$ where $$F_\mu = \mu*m_1g$$ where $$\mu$$ is coefficient of friction. The remaining problem can be solved by substituting for T and plugging the values in.

### Subject: Physics (Electricity and Magnetism)

Calculate the magnitude of the force exerted by two 5m long straight current-carrying wires carrying 1A current in opposite direction kept at a distance of 10cm.

The magnetic field at a distance r from current-carrying wire can be given as $$B = \frac{\mu_0 I_1}{2\pi r}$$ Force on the other wire can be given as $$F = I_2dlB$$ dl is the length of wire. substituting for B, $$F = \frac{\mu_0 I_1I_2dl }{2\pi r}$$ Substituting values, $$F =- \frac{2*10^{-7}(5)}{ (0.1)}$$ $$F =- 10^{-5}N$$ Hence the force exerted by the wires on each other is $$10^{-5}N$$ and the minus sign tells us that the force is attractive.

### Subject: Physics

what is the magnitude of a final velocity when the object hits the ground which is thrown up by making an angle of $$30^{\circ}$$ with horizontal with an initial velocity of 10m/s also find the time of flight.

To find out the final velocity both horizontal and vertical components of the final velocity must be known. In projectile motion, horizontal velocity doesn't change while vertical velocity is affected by gravitational acceleration. those velocities can be given as follows x component of velocity, $$v_x= v_0cos(\theta)$$ $$v_x= 10cos(30)$$ $$v_x=8.66 m/s$$ y component of velocity, $$v_y= v_0sin(\theta)$$ $$v_y= 10sin(30)$$ $$v_y= 5 m/s$$ In projectile motion, after hitting the ground verticle displacement of an object is 0. Hence kinematic equations can be used as follows. $$x=v_it+1/2at^2 $$ as $$v_i=5m/s$$ and $$x = 0$$ above equation becomes $$0=5t+1/2(-9.8)t^2 $$ $$0=5t-4.9t^2 $$ $$0=t(5-4.9t) $$ $$t = 0s$$ and $$t = 1.02s$$ Now, the final horizontal velocity is same as the initial horizontal velocity. To find the final vertical velocity, first, find the maximum high reached. It can be found by using the kinematic equation, $$v_f^2=v_i^2+2ad$$ along y-direction at maximum hight, $$v_f=0$$ and d is a displacement. $$0=v_i^2+2(-9.8)(d)$$ $$0 = 25-19.6d$$ $$25/19.6=d$$ $$d=1.3m$$ Now, final vertical velocity can b found using this maximum distance "d" by using kinematic equation, $$v_f^2=v_i^2t+2ad $$ at the topmost point, $$v_i=0$$ $$v_f^2=-19.6( -1.3)$$ ...........(distance is -ve as object traveled downwards) $$V_f=5 $$ Hence it can be observed that initial and final components of velocities are the same hence magnitude of final velocity is also the same as initial i.e. 10m/s.

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