TS EAMCET 2019 Physics Question Paper with Answer and Solution

(A) Given: mass $m = 2 \,kg$, initial velocity $\vec{u} = (20 \hat{i} + 24 \hat{j}) \,m/s$.
The vertical component of velocity at time $t$ is given by $v_y = u_y - gt$.
At $t = 8 \,s$, $v_y = 24 - (10 \times 8) = 24 - 80 = -56 \,m/s$.
The horizontal component remains constant: $v_x = 20 \,m/s$.
So, final velocity $\vec{v} = (20 \hat{i} - 56 \hat{j}) \,m/s$.
By the Work-Energy Theorem, the change in potential energy $\Delta PE$ is equal to the negative of the change in kinetic energy $\Delta KE$.
$\Delta PE = -\Delta KE = -(KE_f - KE_i) = KE_i - KE_f$.
$\Delta PE = \frac{1}{2} m (u^2 - v^2) = \frac{1}{2} \times 2 \times [(20^2 + 24^2) - (20^2 + (-56)^2)]$.
$\Delta PE = (400 + 576) - (400 + 3136) = 976 - 3536 = -2560 \,J$.
Converting to kilojoules, $\Delta PE = -2.56 \,kJ$.

(D) According to the law of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.
$m_1 u_1 + m_2 u_2 = (m_1 + m_2) v$
Given, mass of boxes $m_1 = m_2 = 3 \,kg$, speed of the moving box $u_1 = 4 \,m/s$, and initial speed of the second box $u_2 = 0$.
Substituting the values:
$3 \times 4 + 3 \times 0 = (3 + 3) v$
$12 = 6v \Rightarrow v = 2 \,m/s$
Thus, the two bodies move together with a velocity of $2 \,m/s$ just after the collision.
Now, applying the law of conservation of energy from the edge of the table to the floor:
$Total Energy_{initial} = Total Energy_{final}$
$KE_{initial} + PE_{initial} = KE_{final} + PE_{final}$
Taking the floor as the reference level $(h=0)$:
$\frac{1}{2} (m_1 + m_2) v^2 + (m_1 + m_2) g h = KE_{final} + 0$
$\frac{1}{2} \times (3 + 3) \times (2)^2 + (3 + 3) \times 10 \times 1 = KE_{final}$
$\frac{1}{2} \times 6 \times 4 + 6 \times 10 = KE_{final}$
$12 + 60 = 72 \,J$
Hence, the kinetic energy just before the boxes strike the floor is $72 \,J$.

(D) Given:
Mass of bullet $m = 1 \ kg$
Initial velocity $u = 2 \ m \ s^{-1}$
Retarding force $F = -0.5/x$
According to the work-energy theorem,the work done by the force equals the change in kinetic energy:
$W = \Delta KE = K_f - K_i$
Since the bullet stops,$K_f = 0$,so $W = -K_i = -\frac{1}{2} m u^2$
$W = -\frac{1}{2} \times 1 \times (2)^2 = -2 \ J$
Also,$W = \int_{x_1}^{x_2} F \ dx = \int_{10-L/2}^{10+L/2} -\frac{0.5}{x} \ dx$
$-0.5 [\ln(x)]_{10-L/2}^{10+L/2} = -2$
$\ln \left( \frac{10+L/2}{10-L/2} \right) = \frac{-2}{-0.5} = 4$
$\frac{10+L/2}{10-L/2} = e^4 = 55$
$10 + L/2 = 55(10 - L/2)$
$10 + L/2 = 550 - 27.5L$
$28L = 540$
$L = 540 / 28 \approx 19.28 \ m$
Rounding to $1$ decimal digit,$L = 19.3 \ m$.

(C) Let the compression of the spring be $x$ (in meters).
According to the law of conservation of mechanical energy,the total potential energy lost by the mass equals the potential energy gained by the spring.
The total height fallen by the mass is $(h + x)$,where $h = 1.2 \ m$.
So,$mg(h + x) = \frac{1}{2} kx^2$.
Substituting the given values: $2 \times 10 \times (1.2 + x) = \frac{1}{2} \times (2 \times 10^4) \times x^2$.
$20(1.2 + x) = 10^4 x^2$.
$24 + 20x = 10000x^2$.
$10000x^2 - 20x - 24 = 0$.
Dividing by $4$: $2500x^2 - 5x - 6 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{5 \pm \sqrt{(-5)^2 - 4(2500)(-6)}}{2(2500)} = \frac{5 \pm \sqrt{25 + 60000}}{5000} = \frac{5 \pm \sqrt{60025}}{5000} = \frac{5 \pm 245}{5000}$.
Since $x > 0$,$x = \frac{250}{5000} = 0.05 \ m$.
Converting to $mm$: $0.05 \ m = 50 \ mm$.

(A) Given,tangential acceleration $a_t = \alpha x^2$.
Tangential force $F = m a_t = m \alpha x^2$.
According to the work-energy theorem,the work done by the tangential force is equal to the change in kinetic energy: $W = \Delta K$.
Assuming the particle starts from rest at $x = 0$,$K = \int_0^x F dx$.
$K = \int_0^x (m \alpha x^2) dx$.
$K = m \alpha \left[ \frac{x^3}{3} \right]_0^x = \frac{m \alpha}{3} x^3$.
Comparing this with the given expression $K = \beta x^c$,we get $\beta = \frac{m \alpha}{3}$ and $c = 3$.

(B) Let $t_A = 2 \ s$ be the time interval for boy $A$ and $t_B = 1 \ s$ be the time interval for boy $B$.
For a ball projected vertically,the time taken to reach the maximum height from a point at height $h$ is $t/2$.
Using $v = u + at$,at the highest point $v = 0$,so $u = g(t/2)$.
The height $h$ of the point from the ground is given by $h = u(t/2) - 1/2 g(t/2)^2 = g(t/2)^2 - 1/2 g(t/2)^2 = 1/2 g(t/2)^2 = 1/8 g t^2$.
For boy $A$: $h_A = 1/8 \times 10 \times (2)^2 = 5 \ m$.
For boy $B$: $h_B = 1/8 \times 10 \times (1)^2 = 1.25 \ m$.
The difference in vertical positions is $h_A - h_B = 5 \ m - 1.25 \ m = 3.75 \ m$.

(A) conservative force is defined as a force where the total work done in moving a particle between two points is independent of the path taken.
It depends solely on the initial and final positions of the particle.
Mathematically,the work done by a conservative force around any closed path is zero.
Therefore,the correct option is $A$.

(C) Given: Mass of the metal chain $m = 2 \,kg$, total length $l = 90 \,cm = 0.9 \,m$.
Length of the hanging part $l' = (90 - 60) \,cm = 30 \,cm = 0.3 \,m$.
The mass of the hanging part $m'$ is proportional to its length: $m' = (l'/l) \times m = (0.3 / 0.9) \times 2 = 2/3 \,kg$.
The center of gravity of the hanging part is at a distance $l_c = l'/2 = 0.3 / 2 = 0.15 \,m$ below the table surface.
The work done to pull the hanging part onto the table is equal to the change in potential energy: $W = m' g l_c$.
Substituting the values: $W = (2/3) \times 10 \times 0.15 = 1 \,J$.