### Shortcut Methods

**Typical Numerical Questions and Shortcuts**

**1. Two Objects Colliding:**

Initial momentums: (p_1 = m_1v_1) and (p_2 = m_2v_2)

Final momentum (after collision): (p_f = (m_1 + m_2)v)

Using conservation of momentum: $$p_1 + p_2 = p_f $$ $$m_1v_1 + m_2v_2 = (m_1 + m_2)v$$ $$v = \frac{m_1v_1 + m_2v_2}{m_1 + m_2}$$

**2. Ball Colliding with a Wall**

Impulse: (J = \Delta p)

Initial momentum: (p_i = mv)

Final momentum: (p_f = -mv) (negative sign due to the change in direction)

Impulse: $$J = p_f - p_i = -2mv$$

**3. Rocket Expelling Gas**

Initial momentum of rocket and gas: (p_i = (M + m)v)

Momentum of the gas after expulsion: (p_g = mu) (negative sign since gas moves in the opposite direction)

Momentum of the rocket after expulsion: (p_r = (M + m)v’ ) (v’ = v + \frac{-mu}{M + m})

**4. Potential Energy in a Spring**:

$$PE = \frac{1}{2}kd^2$$

where (k) is the spring constant and (d) is the compression or extension.

**5. Work Done by a Force:**

$$W = Fd\cos\theta$$ where (F) is the force, (d) is the displacement, and (\theta) is the angle between the force and the displacement.

**6. Speed of Pendulum Bob at Lowest Point:**

Initial potential energy: (PE = mgh)

Final kinetic energy at the lowest point: (KE = \frac{1}{2}mv^2)

Using conservation of energy:

$$PE = KE$$ $$mgh = \frac{1}{2}mv^2$$ $$ v= \sqrt{2gh}$$ where (g) is the acceleration due to gravity and (h) is the initial height.

**7. Collision of Two Cars:**

Using conservation of momentum:

Total momentum before collision: $$P = m_1v_1 + m_2(0)$$

Total momentum after collision: $$P = (m_1 + m_2) v_f$$

Equate the two: $$m_1v_1 = (m_1 + m_2) v_f$$ $$ v_f = \frac{m_1v_1}{m_1 + m_2}$$ where (v_f) is the final velocity.

**8. Recoil Velocity of a Gun:**

Using conservation of momentum:

Total momentum before firing: $$P_i = 0$$

Total momentum after firing: $$P_f = Mv_g + mv$$

Equate the two: $$0 = Mv_g + mv$$ $$v_g = - \frac{m}{M} v$$

where (M) is the mass of the gun and (v_g) is the recoil velocity of the gun.

**9. Maximum Height Reached by a Vertically Thrown Ball**

At the highest point, kinetic energy is zero, and potential energy is maximum:

$$KE_i + PE_i = PE_{max}$$ $$\frac{1}{2}mv^2 + 0 = mgh$$ $$h = \frac{v^2}{2g}$$

where (g) is the acceleration due to gravity and (v) is the initial velocity.

**10. Maximum Height Reached by a Rocket:**

Initial kinetic energy: (KE = \frac{1}{2}mv^2)

Final potential energy at the highest point: (PE = mgh)

Using conservation of energy: $$KE = PE$$ $$\frac{1}{2}mv^2 = mgh$$ $$h = \frac{v^2}{2g}$$

where (g) is the acceleration due to gravity.