Two bodies of different masses \(m_a\) and \(m_b\) are dropped from two different heights, viz, an and b. The ratio of time taken by the two to drop through these distances is :
The time taken to fall through height \(h\) from rest under gravity is \(t = \sqrt{\frac{2h}{g}}\). Since \(t \propto \sqrt{h}\), the ratio of times for heights \(a\) and \(b\) is \(\sqrt{a} : \sqrt{b}\).
A particle moves along the parabolic path \(y = ax^2\) in such a way that x-component of velocity remains constant, say c. The acceleration of the particle is :
Given \(v_x = \frac{dx}{dt} = c\) (constant), hence \(a_x = 0\). Differentiating \(y = ax^2\) gives \(v_y = 2ax\frac{dx}{dt} = 2axc\). Differentiating again, \(a_y = 2ac\frac{dx}{dt} = 2ac^2\). Therefore, total acceleration vector is \(a = 2ac^2\hat{j}\).
If a body A of mass M is thrown with velocity \(v\) at an angle \(30^\circ\) to the horizontal and another body B of same mass is thrown with same speed at an angle of \(60^\circ\) to the horizontal, the ratio of range of A and B will be :
Horizontal range is given by \(R = \frac{u^2 \sin(2\theta)}{g}\). For complementary projection angles \(\theta\) and \(90^\circ - \theta\) (like \(30^\circ\) and \(60^\circ\)), the horizontal ranges are equal. Thus, the ratio of range is \(1 : 1\).
At what angle to the horizontal should an object be projected so that the maximum height reached is equal to the horizontal range?
Maximum height \(H = \frac{u^2\sin^2\theta}{2g}\) and range \(R = \frac{2u^2\sin\theta\cos\theta}{g}\). Equating the two yields \(\frac{\sin^2\theta}{2} = 2\sin\theta\cos\theta\), which simplifies to \(\tan\theta = 4\).
A body of mass \( m \) is projected along a rough inclined plane (having an angle of inclination with horizontal \( \theta \), equal to angle of repose) with a velocity \( v \). It travels up a maximum distance \( s \) before it comes to a halt. Then \( v \) is:
Since the inclination equals the angle of repose, \( \mu = \tan\theta \). The acceleration down the incline during upward motion is \( a = g \sin\theta + \mu g \cos\theta = 2g \sin\theta \). Using \( v^2 = 2as \), we get \( v = 2\sqrt{gs \sin\theta} \).
At what angle to the horizontal should an object be projected so that the maximum height reached is equal to half of the horizontal range?
We require \(H = \frac{R}{2}\). Using the formulae \(H = \frac{u^2 \sin^2 \theta}{2g}\) and \(R = \frac{u^2 \sin 2\theta}{g}\), we get \(\frac{u^2 \sin^2 \theta}{2g} = \frac{u^2 \sin\theta\cos\theta}{g} ⇒ \tan\theta = 2 ⇒ \theta = \tan^{-1}(2)\).
A body moves from A to B with a constant speed of \(20\text{ ms}^{-1}\) and returns from B to A with a constant speed of \(40\text{ ms}^{-1}\). The average speed of the body for the whole journey is
For equal distances covered in two halves of a journey, average speed is \(v_{\text{avg}} = \frac{2v_1v_2}{v_1+v_2}\). Here, \(v_{\text{avg}} = \frac{2 \times 20 \times 40}{20+40} = \frac{80}{3}\text{ ms}^{-1}\).
A stone dropped from the top of a tower travels \(\frac{5}{9}\) th of the height of tower during the last second of fall. Height of the tower is: (take \(g = 10\text{ m/s}^2\))
Distance in last second is \(h_{\text{last}} = \frac{5}{9}H ⇒ \text{Distance in } (t-1) \text{ seconds is } \frac{4}{9}H\). Therefore, \(\frac{\frac{1}{2}g(t-1)^2}{\frac{1}{2}gt^2} = \frac{4}{9} ⇒ \frac{t-1}{t} = \frac{2}{3} ⇒ t = 3\text{ s}\). Height \(H = \frac{1}{2}gt^2 = \frac{1}{2} \times 10 \times 9 = 45\text{ m}\).
If velocity of a car increases uniformly from \(20\text{ m/sec}\) to \(60\text{ m/sec}\) in a time interval of 5 seconds. Then distance travelled during this interval is:
Since acceleration is uniform, the distance is given by the formula \(s = \frac{u+v}{2} \times t\). Substituting \(u=20\text{ m/s}\), \(v=60\text{ m/s}\), and \(t=5\text{ s}\), we get \(s = \frac{20+60}{2} \times 5 = 200\text{ m}\).
Between two stations a train first accelerates uniformly, then moves with uniform speed and finally retards uniformly. If the ratios of the time taken for acceleration, uniform speed and retarded motions are \(1 : 8 : 1\) and the maximum speed of the train is \(60\text{ km/hr}\), the average speed of the train over the whole journey is:
Let the time intervals be \(t\), \(8t\), and \(t\). Total time is \(10t\). Total distance is the area under the v-t graph: \(S = \frac{1}{2} (8t + 10t) v_{\text{max}} = 9t v_{\text{max}}\). Average speed is \(v_{\text{avg}} = \frac{9t v_{\text{max}}}{10t} = 0.9 v_{\text{max}} = 0.9 \times 60 = 54\text{ km/hr}\).