Question

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 forces during the crash.36.9° from the x-axis

Answer 1

Refer to the diagram shown below.

Define the (x,y) plane as the horizontal plane of the floor.
There was no momentum in the (x,y) plane before the plate hit the floor.

Let the velocity components in the (x) and (y) directions of the 100 g mass be Vx and Vy respectively, and that the resultant velocity, V, makes an angle θ below the negative x-axis as shown.

Because momentum is conserved, therefore
100*Vx + 320*2 = 0
100Vx = -640
Vx = -6.4 m/s

100Vy + 355*1.5 = 0
100Vy = -532.5
Vy = -5.325 m/s

V = √[(-6.4)² + (-5.325)²] = 8.33 m/s
θ = tan⁻¹ (-5.325/-6.4) = 39.8°

Answer:
The direction is 39.8° below the negative x-axis
The speed is 8.33 m/s

Answer 2

The direction of third piece is $\boxed{{{39.8}^ \circ }}$ form the positive $x$  axis.

Explanation:

Three pieces of plate are formed after falling vertically to floor and slides along the floor and goes in three directions.

The first piece of $320{\text{ g}}$ moves on the x-axis at the speed of $2{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$ and the other piece of $355{\text{ g}}$  moves on the y-axis at the speed of $1.5{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$.

Our aim is to find the direction of third piece that weighs $100{\text{ g}}$ .

Consider the horizontal plane of floor be x-y plane. Before the collision of plate to the floor there was zero momentum.

Suppose the components of velocity of third piece be ${v_x}$  and ${v_y}$ , and the resultant velocity $v$  makes an angle of ${{\alpha }}$  under the negative x-axis as shown in Figure 1.

Due to conservation of momentum the equation can be written as,

$\begin{aligned}100{v_x}+2\left({320}\right)&=0\\100{v_x}&=-640\\{v_x}&=-6.4{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{aligned}$

The other equation from conservation of momentum can be written as,

$\begin{aligned}100{v_y}+1.5\left({355}\right)&=0\\100{v_y}&=-532.5\\{v_y}&=-5.325{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{aligned}$

The velocity $v$  by which the third piece moves is calculated as,

$\begin{aligned}v&=\sqrt {v_x^2+v_y^2}\\&=\sqrt{{{\left({-6.4}\right)}^2}+{{\left({-5.325}\right)}^2}}\\&=8.3256{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{aligned}$

The direction ${{\alpha }}$  that $v$  with the vertical direction is calculated as,

$\begin{aligned}tan\alpha&=\frac{v_y}{v_x}\\&=\frac{-5.325}{6.4}\\&=0.832\end{aligned}$

The value of ${\alpha }$  is calculated as,

$\begin{aligned}\alpha&=tan^{-1}(0.832)\\&=39.76^{\circ}\\&\approx 39.8^{\circ}\end{aligned}$

Therefore, the value of angle ${\alpha }$  is ${39.8^ \circ }$ .

Thus, the direction of third piece is $\boxed{{{39.8}^ \circ }}$  form the positive $x$  axis.

Learn More:

1. Velocity and Momentum brainly.com/question/11896510

2. Linear momentum brainly.com/question/11947870

3. Motion and velocity brainly.com/question/6955558

Answer Details:

Grade: High School

Subject: Physics

Chapter: Momentum

Keywords:

Plate, vertically, floor, breaks, three, pieces, slide, impact, 320 g, 2 m/s, 355 g, 1.50 m/s, 100 g, direction, motion, horizontal, force, x-axis, y-axis, momentum, falls, collision.