Question

Two blocks of clay slide on a level, frictionless plane. Block A, having a mass of 2.00 kg and traveling due south, collides with block B, having a mass of 3.00 kg and traveling 25.0 degrees west of due north. They stick together after the collision, and the resulting blob travels with a final speed of 9.89 m/s in a direction 39.9 degrees west of due north. Calculate the initial speeds of blocks A and B.

Answer 1

Answer:

$v_1 =$-15.03m/s

$v_2 =$25.01 m/s

Explanation:

Using the conservation of the linear momentum:

$L_i = L_f$

So, for axis x:

$m_1v_{1x}+m_2v_{2x} = m_sv_{sx}$

where $m_1$ is the mass of the block A, $v_{1x}$ is the velocity in x of the block A, $m_2$ is the mass of the block B, $v_{2x}$ is the velocity in x of the block B, $m_s$ the mass of the block A and B together and $v_{sx}$ the velocity in x of both blocks after the collition.

Replacing values:

$(2kg)(0)+(3kg)v_2sin(25) = (5kg)(9.89m/s)sin(39.9)$

$(3kg)v_2sin(25) = (5kg)(9.89m/s)sin(39.9)$

Solving for $v_2$:

$v_2 =$25.01 m/s

At the same way, for axis y:

$m_1v_{1y}+m_2v_{2y} = m_sv_{sy}$

where $m_1$ is the mass of the block A, $v_{1y}$ is the velocity in y of the block A, $m_2$ is the mass of the block B, $v_{2y}$ is the velocity in y of the block B, $m_s$ the mass of the block A and B together and $v_{sy}$ the velocity in y of both blocks after the collition.

$(2kg)v_1+(3kg)(25.01m/s)cos(25) = (5kg)(9.89m/s)cos(39.9)$

solving for $v_1$:

$v_1 =$-15.03m/s

it is negative because we set the north as positive.