# A plate falls vertically to the floor and breaks up into three pieces, which slide along the floor. Immediately after the impact, a 320-g piece moves along the x-axis with a speed of 2.00 m/s and a 355-g piece moves along the y-axis with a speed of 1.50 m/s. The third piece has a mass of 100 g. In what direction does the third piece move?You can neglect any horizontal during the crash.

This problem involves the momentum-impulse theorem, also known as the momentum form of the second Newton's Law:

`vecp_f - vecp_i = vecF Deltat`

Here, `vecp_i` and `vecp_f` are initial and final momenta of the plate, respectively, and `vecF` is the force acting on the plate during its collision with the floor, which lasted some time `Delta t`. Note that initial momentum is the momentum of the whole plate right before it hit the floor, and the final momentum is the sum of the momenta of all the pieces immediately after the plate broke.

To use the momentum-impulse theorem, we need to bring it from the vector form to the component form as follows:

`p_(fx) - p_(ix) = F_x Delta t`

`p_(fy) - p_(iy) = F_yDeltat`

`p_(fz) - p_(iz) = F_zDeltat`

The last equation (for z-component, along the vertical line) is not needed because we do not know either vertical component of the initial momentum nor the vertical force acting on the plate from the floor. We do know that `p_(fz) = 0`.

For the x- and y-components, `p_(ix) = 0` and `p_(iy) = 0` because initially the plate is moving in only vertical direction. Also, `F_x = 0` and `F_y = 0` because the force on the plate from the floor is normal (perpendicular to the floor), so it has only z-component. So, `p_(fx) = 0` and `p_(fy) = 0`.

The x-component of the total momentum is the sum of the x-components of the momenta of the three pieces as follows:

`p_(fx) = p_(1x) + p_(2x) + p_(3x) = 0`

We are given that `p_1x = m_1v_1 = 0.32*2 kg*m/s = 0.64 kg*m/s` , `p_(2x) = 0` and `p_(3x) = m_3v_(3x) ` , where `v_(3x)` is unknown.

Similarly,

`p_(fy) = p_(1y) + p_(2y) + p_(3y) = 0`. Here, `p_(1y) = 0` , `p_(2y) = m_2v_(2y) = 0.355*1.5 kg*m/s = 0.5325 kg*m/s ` and `p_(3y) = m_3v_(3y)` , where `v_(3y) ` is unknown.

Plugging everything in, we have the following:

`0.64 + 0.1v_(3x) = 0`

and

`0.5325 + 0.1v_(3y) = 0`

From here, we can get to the following:

`v_(3x) = -6.4 m/s` and `v_(3y) = -5.325 m/s`

From the x- and y-components of the velocity of the third piece vector, we can find its magnitude (speed of the third piece) and the direction of its motion:

`v_3 = sqrt(v_(3x) ^2 + v_(3y)^2) = 8.33 m/s` (using the three significant digits, since the given values are given with the three significant digits.)

Since the x-and y-components of the velocity are negative, the velocity vector is in the fourth quadrant and it makes an acute angle with the negative x-axis. This angle is given by the following:

`tan theta = v_y/(v_x)= (-5.325)/(-6.4) = 0.832`

and `theta = 39.8` degrees.

So the third piece moves with the speed of 8.33 m/s making an angle of 39.8 degrees with the negative x-axis.

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Considering the system of coordinates with (x,y) axis in the floor plane, the fall of the plate is initially directed towards the negative direction of z axis. Touching the floor the force on the plate is normal on it, thus on the positive direction of z axis and cancels the initial (-z) direction momentum.  Since there is no force on the plate on the (x,y) plane, the total impulse on the (x,y) plane need to be the same before and after the collision. Therefore written as vectors one has

`m_1*v_x +m_2*v_y +m_3*v_3 =0`

`|v_3| = sqrt((m_1*v_x)^2+ (m_2*v_y)^2)/m_3`

`|v_3| =sqrt((0.32*2)^2 +(0.355*1.5)^2)/0.1 =8.326 m/s`

The angle that `v_3` is making with the positive direction of the x axis is

`theta = 180 +arctan ((m_2*v_y)/(m_1*v_x)) = 180+arctan((0.355*1.5)/(0.32*2)) =180+39.76 =219.76 degree`

The figure is below.

Answer: the third piece moves in the (x,y) plane making an agle of 219.76 degree with the positive direction of x axis.