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 (\(10\text{ m/s}^2\)).
A particle moving with uniform acceleration crosses two points A and B present in a straight line with speed 10 m/s and 20 m/s respectively, the speed of particle at mid-point of A and B will be
1. \[5\sqrt{10}\text{ m/s}\]
2. \[10\sqrt{5}\text{ m/s}\]
3. \[ \sqrt{5}\text{ m/s}\]
4. \[\sqrt{10}\text{ m/s}\]
View Answer
Formula: Under uniform acceleration, the midpoint velocity is \(v_{mid} = \sqrt{\frac{v_1^2 + v_2^2}{2}}\). Thus, \(v_{mid} = \sqrt{\frac{100 + 400}{2}} = 5\sqrt{10}\text{ m/s}\).
Assertion (A): A particle with constant acceleration always moves along a straight line.
Reason (R): A particle with constant acceleration will not change direction of motion.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
A projectile experiences constant acceleration (g) but follows a parabolic path. Also, a ball thrown vertically upwards under gravity has constant acceleration but reverses its direction of motion, making both statements false.
Assertion (A): If initial velocity is negative and acceleration is positive then motion is retarded (initially).
Reason (R): If initial velocity is negative but acceleration is positive then displacement of a particle can never be positive.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
When velocity and acceleration have opposite signs, the speed decreases (retardation). After stopping, the positive acceleration will move the particle in the positive direction, which can result in positive displacement, so R is false.
Assertion (A): A particle with constant acceleration always moves along a straight line.
Reason (R): A particle with constant acceleration will not change direction of motion.
1. (1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. (2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (3) (A) is true but (R) is false
4. (4) Both (A) and (R) are false
View Answer
Assertion (A) is false. Projectile motion has constant acceleration (\(\vec{g}\)) but follows a parabolic path, not a straight line.
Reason (R) is false. A particle can change direction even with constant acceleration (e.g., projectile motion, or an object slowing down and reversing direction).
Thus, both (A) and (R) are false.