A particle is projected from horizontal ground with speed \(50\text{ ms}^{-1}\) at \(53^\circ\) with horizontal. Find time after which velocity of particle will be at \(45^\circ\) with horizontal for the second time.
1. \(1\text{ sec}\)
2. \(3\text{ sec}\)
3. \(5\text{ sec}\)
4. \(7\text{ sec}\)
View Answer
Horizontal speed is \(u_x = u\cos(53^\circ) = 30\text{ m/s}\), and vertical speed is \(u_y = u\sin(53^\circ) = 40\text{ m/s}\). At \(45^\circ\) on the way down, \(v_y = -v_x = -30\text{ m/s}\). Using \(v_y = u_y - gt\), we get \(-30 = 40 - 10t\), which yields \(t = 7\text{ s}\).
A point moves linearly such that \(\frac{dv}{dt} = -\alpha v^{1/2}\), where \(\alpha\) is a positive constant. At start, \(v = v_0\). Find time taken by particle to stop?
1. \(\frac{\sqrt{v_0}}{\alpha}\)
2. \(\frac{v_0}{\alpha}\)
3. \(\frac{2\sqrt{v_0}}{\alpha}\)
4. \(\alpha\sqrt{v_0}\)
View Answer
Integrating \(v^{-1/2} dv = -\alpha dt\) from \(v_0\) to \(0\) gives \(2\sqrt{v_0} = \alpha t\), so the time taken to stop is \(t = \frac{2\sqrt{v_0}}{\alpha}\).
A paratrooper jumps from a height \(H\). The parachute can provide a uniform deceleration of \(2\text{ ms}^{-2}\). The height above the ground at which the parachute should be opened so that he touches ground with zero speed is (take \(g = 10\text{ ms}^{-2}\)):
1. \(\frac{H}{6}\)
2. \(\frac{4H}{5}\)
3. \(\frac{5H}{6}\)
4. \(\frac{6H}{7}\)
View Answer
Let \(h\) be free fall and \(y\) be decelerating height, so \(H = h + y\). Speed before parachute opens: \(v^2 = 2gh = 20h\). Deceleration phase: \(0 = v^2 - 2ay = v^2 - 4y\), which gives \(20h = 4y \Rightarrow y = 5h\). Since \(H = 6h\), we find \(y = \frac{5H}{6}\).
When a motorcycle moving with a uniform speed \(11\text{ m/s}\) is at a distance \(24\text{ m}\) from a car, the car starts from rest and moves with a uniform acceleration \(2\text{ m/s}^2\) away from the motorcycle. If the car begins motion at \(t = 0\), time at which the motorcycle will overtake the car is \(t = \):
1. \(8\text{ sec}\)
2. \(6\text{ sec}\)
3. \(3\text{ sec}\)
4. \(1.5\text{ sec}\)
View Answer
Distance equation for meeting: \(11t = 24 + \frac{1}{2}(2)t^2 \Rightarrow t^2 - 11t + 24 = 0\). Solving this quadratic equation gives \(t = 3\text{ s}\) and \(t = 8\text{ s}\). The first overtake occurs at \(t = 3\text{ s}\).
A particle moves in a straight line with velocity \(v = (4t – 2)\text{ m/s}\). If the particle starts from \(x = 0\), then the position of particle at which it momentarily comes to rest is
1. \(x = -0.5\text{ m}\)
2. \(x = 5\text{ m}\)
3. \(x = 2\text{ m}\)
4. \(x = 4\text{ m}\)
View Answer
The particle comes to rest when \(v = 0\), giving \(4t - 2 = 0 ⇒ t = 0.5\text{ s}\). Integrating velocity to find position, \(x = int_0^t (4t - 2)dt = 2t^2 - 2t\). At \(t = 0.5\text{ s}\), \(x = 2(0.5)^2 - 2(0.5) = -0.5\text{ m}\).
A car starts from rest and moves with constant acceleration \(a\). The ratio of distances covered in the first second to the distance covered in third second is
1. \(1 : 2\)
2. \(2 : 1\)
3. \(3 : 1\)
4. \(1 : 5\)
View Answer
The distance covered in the \(n\)-th second from rest is given by \(S_n = frac{a}{2}(2n - 1)\). For the first second (\(n=1\)), \(S_1 = frac{a}{2}\), and for the third second (\(n=3\)), \(S_3 = frac{5a}{2}\). The ratio of \(S_1 : S_3\) is \(1 : 5\).
A particle is moving in vertical plane (x-y plane) such that its trajectory is given by the equation, \(y = x – \frac{x^2}{80}\), where \(x\) & \(y\) are in metre. For this particle match column-I with column-II and tick the correct option.
**Column I**:
A. Horizontal range (in m)
B. Angle of projection with horizontal (degree)
C. Maximum height gained by particle (m)
D. Speed of projection (\(\text{m s}^{-1}\))
**Column II**:
(P) 45
(Q) 80
(R) 20
(S) \(20\sqrt{2}\)
1. A(S), B(R), C(Q), D(P)
2. A(P), B(Q), C(S), D(R)
3. A(Q), B(P), C(S), D(R)
4. A(Q), B(P), C(R), D(S)
View Answer
Comparing the trajectory to \(y = xtan\theta - \frac{gx^2}{2u^2cos^2\theta}\), we find \(tan\theta = 1⇒ \theta = 45^\circ\). Range at \(y = 0\) is \(x = 80\text{ m}\). Maximum height is \(H = \frac{Rtan\theta}{4} = 20\text{ m}\), and the initial speed \(u = 20\sqrt{2}\text{ m/s}\). Hence, A-Q, B-P, C-R, D-S.
A small block slides down on a smooth inclined plane, starting from rest at time \(t = 0\). Let \(S_n\) be the distance travelled by the block in the interval \(t = n – 1\) to \(t = n\). Then, the ratio \(\frac{S_n}{S_{n+1}}\) is
1. \(\frac{2n}{2n-1}\)
2. \(\frac{2n-1}{2n}\)
3. \(\frac{2n-1}{2n+1}\)
4. \(\frac{2n+1}{2n-1}\)
View Answer
The distance in the \(n^{\text{th}}\) second is \(S_n = u + \frac{a}{2}(2n-1)\). Since \(u=0\), \(S_n \propto 2n-1\). Thus, \(\frac{S_n}{S_{n+1}} = \frac{2n-1}{2(n+1)-1} = \frac{2n-1}{2n+1}\).
A car starts from rest and accelerates at \(5\text{ m/s}^2\). At \(t = 4\text{ s}\), a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at \(t = 6\text{ s}\)? (Take \(g = 10\text{ m/s}^2\))
1. \(20\sqrt{2} m/s, 10 m/s^2\)
2. \(20 m/s, 5 m/s^2\)
3. \(20 m/s, 0\)
4. \(20\sqrt{2} m/s, 0\)
View Answer
At \(t = 4\text{ s}\), the horizontal velocity is \(v_x = u + at = 0 + 5 \times 4 = 20\text{ m/s}\) and remains constant. In the vertical direction, the ball accelerates due to gravity for \(2\text{ s}\) (from \(t = 4\text{ s}\) to \(6\text{ s}\)), so \(v_y = gt = 10 \times 2 = 20\text{ m/s}\). The final velocity is \(v = \sqrt{v_x^2 + v_y^2} = 20\sqrt{2}\text{ m/s}\), and the acceleration is only due to gravity (\(10\text{ m/s}^2\)).