An object of mass \(m_1\) collides with another object of mass \(m_2\), which is at rest. After the collision the objects move with equal speeds in opposite direction. The ratio of the masses \(m_2 : m_1\) is:

Question

Three bodies A, B and C have equal kinetic energies and their masses are 400 g, 1.2 kg and 1.6 kg respectively. The ratio of their linear momenta is :

Answer 1

To determine the ratio of the masses \( m_2 : m_1 \), we need to analyze the collision described in the problem using the principle of conservation of linear momentum. The key points to note from the problem are:

An object of mass \( m_1 \) is moving and collides with another object of mass \( m_2 \), which is initially at rest.
After the collision, both objects move with equal speeds in opposite directions.

Since the problem involves a collision, the principle of conservation of momentum applies. Before the collision, the total momentum of the system is only due to the object of mass \( m_1 \) since \( m_2 \) is at rest. Let's denote the initial velocity of \( m_1 \) as \( v_1 \).

p_{\text{initial}} = m_1 \cdot v_1 + m_2 \cdot 0 = m_1 \cdot v_1

After the collision, both objects move with equal speed \( v \) but in opposite directions; let's denote the velocity of \( m_1 \) as \( v \) and \( m_2 \) as \( -v \). Therefore, the total momentum after the collision is:

p_{\text{final}} = m_1 \cdot (-v) + m_2 \cdot v = -m_1 \cdot v + m_2 \cdot v

According to the principle of conservation of momentum:

m_1 \cdot v_1 = -m_1 \cdot v + m_2 \cdot v

Since the objects move with equal speeds \( v \) after the collision, we can simplify this equation by factoring out \( v \):

m_1 \cdot v_1 = v(m_2 - m_1)

As the objects move with equal speeds, we also know that momentum should be conserved such that:

m_1 = 3m_2

To achieve this condition after equal speeds in opposite directions, we must have the ratio:

m_2 : m_1 = 3 : 1

Thus, the correct answer is that the ratio of the masses \( m_2 : m_1 \) is 3 : 1.

Answer 2