Two particles having masses 4 g and 16 g respectively are moving with equal kinetic energies. The ratio of the magnitudes of their linear momentum is n : 2. The value of n will be ___________.

Answer

To solve the problem, let's denote the masses and velocities of the two particles:

- Mass of particle A (m_A) = 4 g = 0.004 kg (since we need to convert grams to kilograms for standard SI units)
- Mass of particle B (m_B) = 16 g = 0.016 kg
- Velocity of particle A (v_A)
- Velocity of particle B (v_B)

Given that both particles have equal kinetic energies, we can express the kinetic energy (KE) for both particles as follows:

1. Kinetic energy of particle A:
\[ KE_A = \frac{1}{2} m_A v_A^2 \]

2. Kinetic energy of particle B:
\[ KE_B = \frac{1}{2} m_B v_B^2 \]

Since the kinetic energies are equal, we can set these two equations equal to each other:
\[ \frac{1}{2} m_A v_A^2 = \frac{1}{2} m_B v_B^2 \]

We can simplify this equation by canceling out \(\frac{1}{2}\):
\[ m_A v_A^2 = m_B v_B^2 \]

Now, substituting the values of \(m_A\) and \(m_B\):
\[ 0.004 v_A^2 = 0.016 v_B^2 \]

Dividing both sides by 0.004:
\[ v_A^2 = 4 v_B^2 \]

Taking the square root of both sides:
\[ v_A = 2 v_B \]

Next, we need to find the linear momentum (p) of both particles:
1. Linear momentum of particle A:
\[ p_A = m_A v_A = 0.004 v_A \]

2. Linear momentum of particle B:
\[ p_B = m_B v_B = 0.016 v_B \]

Now, substituting \(v_A = 2 v_B\) into the equation for \(p_A\):
\[ p_A = 0.004 (2 v_B) = 0.008 v_B \]

Now we can find the ratio of the magnitudes of their linear momentum:
\[ \frac{p_A}{p_B} = \frac{0.008 v_B}{0.016 v_B} = \frac{0.008}{0.016} = \frac{1}{2} \]

This means the ratio of the magnitudes of their linear momentum is:
\[ p_A : p_B = 1 : 2 \]

According to the problem, this ratio is given as \(n : 2\). Therefore, we can conclude that:
\[ n = 1 \]

Thus, the value of \(n\) will be **1**.