A body of mass 4.2kg moving with velocity 10ms-1 due east, hits a stationary body of mass 2.8kg. If they stick together after collision and move with veloci...
A body of mass 4.2kg moving with velocity 10ms-1 due east, hits a stationary body of mass 2.8kg. If they stick together after collision and move with velocity V due east, calculate the value of V
Answer Details
In this problem, we can use the principle of conservation of momentum, which states that the total momentum of a system remains constant if no external forces act on it.
The initial momentum of the system before the collision is given by:
p1 = m1 * v1
where m1 is the mass of the first body (4.2 kg) and v1 is its velocity (10 m/s).
p1 = 4.2 kg * 10 m/s
p1 = 42 kg m/s
The second body is stationary, so its initial momentum is zero:
p2 = 0 kg m/s
The total initial momentum of the system is therefore:
p_initial = p1 + p2
p_initial = 42 kg m/s
After the collision, the two bodies stick together and move with a common velocity V. The total mass of the system is the sum of the masses of the two bodies:
m_total = m1 + m2
m_total = 4.2 kg + 2.8 kg
m_total = 7 kg
The final momentum of the system is:
p_final = m_total * V
According to the principle of conservation of momentum, the total initial momentum of the system is equal to the total final momentum of the system:
p_initial = p_final
Substituting the values we have found, we get:
42 kg m/s = 7 kg * V
Solving for V, we get:
V = 42 kg m/s / 7 kg
V = 6 m/s
Therefore, the value of V after the collision is 6 m/s due east.
Thus, the correct answer is "6ms^-1".