Shortcut Methods

Projectile Motion:#

Given:

• Angle of projection (θ) = 30°
• Initial velocity (u) = 20 m/s

To find:

• Maximum height reached (H)
• Range of the projectile (R)

Solution:

• Maximum height (H): Using the formula: H = (u² * sin²θ)/(2g) where g = 9.8 m/s² (acceleration due to gravity) H = [(20)² * (sin 30)²]/(2 * 9.8) H ≈ 8.06 m

• Range (R): Using the formula: R = (u² * sin 2θ)/g R = [(20)² * sin 60]/9.8 R ≈ 34.76 m

2. Uniform Circular Motion:

Given:

• Radius of the circular path (r) = 10 m
• Speed of the particle (v) = 5 m/s

To find:

• Period of the motion (T)
• Frequency of the motion (f)

Solution:

• Period (T): Using the formula: T = 2πr/v where π ≈ 3.14 T = 2π * 10 / 5 T ≈ 12.57 s

• Frequency (f): Frequency (f) is the reciprocal of the period (T). f = 1/T f ≈ 1/12.57 f ≈ 0.0796 Hz

3. Relative Motion:

Given:

• River’s speed (v_r) = 2 m/s (towards the east)
• Boat’s speed relative to water (v_bw) = 5 m/s (downstream)

To find:

• Boat’s velocity with respect to the ground (v_bg)

Solution: Since the river is flowing eastward and the boat is moving downstream (in the same direction), we simply add the velocities. V_bg = v_bw + v_r .**

v_bg = 5 m/s (downstream) + 2 m/s (eastward) = 5i + 2j m/s. Therefore, the boat’s velocity with respect to the ground is (5i + 2j) m/s

4. Work-Energy Theorem:

Given:

• Force applied (F) = 10 N
• Distance through which the object is moved (d) = 5 m

To find:

• Work done by the force (W)

Solution: Using the formula: W = F * d W = 10 N * 5 m W = 50 J

5. Collisions:

Given:

• Mass of the first ball (m1) = 10 kg
• Initial velocity of the first ball (u1) = v m/s
• Mass of the second ball (m2) = 5 kg
• Initial velocity of the second ball (u2) = 0 m/s (at rest)
• Velocity loss of the first ball after collision = 25% of u1

To find:

• Velocities of both balls after the collision (v1 and v2)

Solution: Using the law of conservation of momentum: m1 * u1 + m2 * u2 = m1 * v1 + m2 * v2 10 * v + 5 * 0 = 10 * (0.75v) + 5 * v2 7.5v = 5v2

Rearranging the equation, we get: v2 = (1.5v)/5 v2 = 0.3v

Therefore, the final velocities of the 10 kg ball and the 5 kg ball after the collision are 0.75v and 0.3v, respectively.