Mass A hits Mass H initially at rest After the collision A c

Mass A hits Mass H (initially at rest). After the collision. A continues at with a velocity of V_af = 5 at an angle of theta = 53degree. What is the final velocity of B (magnitude and direction)? Masses and initial velocity: m_A = 2 kg m_theta = 4 kg v_A = 6m/sec v_B = 0

Solution

Here as no external force acts on the system, we will use the principle of conservation of momentum, to determine the velocity of the partcile B.

Now, as the particle A starts to move at an angle of 53 degrees and speed of 5 m/s

This momenetum of A will have a vertical as well as horizontal components. Now the initial momentum along the vertical direction was zero,

So we get: 2(5) Sin53 = 4 VSin

or, VSin = 1.9966

Also, the along the horizontal direction we can write:2(6) = 10Cos53 + VCos

or, VCos = 5.982

Therefore = 18.457 which is the required expression.

Putthing this in the equation above, we will get: V Sin = V0.3166 = 1.9966

or, V = 6.3064 is the required speed of the block A