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

`v_1 =`-15.03m/s

`v_2 =`25.01 m/s

Step-by-step 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.