Question

A dinner plate falls vertically to the floor and breaks up into three pieces, which slide horizontally along the floor. immediately after the impact, a 200-g piece moves along the +x-axis with a speed of 2.00 m/s, a 235-g piece moves along the +y-axis with a speed of 1.50 m/s. the third piece has a mass of 100 g. what is the speed of the third piece?

Answer 1

We'll use the momentum-impulse theorem. The x-component of the total momentum in that direction is given by p_(f) = p_(1) + p_(2) + p_(3) = 0.
  So p_(1x) = m1v1 = 0.2 * 2 = 0.4 Also p_(2x) = m2v2 = 0 and p_(3x) = m3v3 = 0.1 *v3 where v3 is unknown speed and m3 is the mass of the third particle with the unknown speed
 Similarly, the 235g particle, y-component of the total momentum in that direction is given by p_(fy) = p_(1y) + p_(2y) + p_(3y) = 0.
 So p_(1y) = 0, p_(2y) = m2v2 = 0.235 * 1.5 = 0.3525 and p_(3y) = m3v3 = 0.1 * v3 where m3 is third particle mass.
  So p_(fx) = p_(1x) + p_(2x) + p_(3x) = 0.4 + 0.1v3; v3 = 0.4/-0.1 = - 4
 Also p_(fy) = 0.3525 + 0.1v3; v3 = - 0.3525/0.1 = -3.525
  So v_3x = -4 and v_3y = 3.525.
 The speed is their resultant = âš (-4)^2 + (-3.525)^2 = 5.335

Answer 2

The speed of third piece that weighs $100{\text{ g}}$  is $\boxed{5.33{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}$ .

Explanation:

Three pieces of dinner plate after falling vertically to floor and slides horizontally along the floor and goes in three different directions.

The first piece of $200{\text{ g}}$  moves along the x-axis by the speed of $2{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$  and the other piece of $235{\text{ g}}$  moves on the y-axis by 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 speed of third piece that weighs $100{\text{ g}}$ .

Let the horizontal plane of floor be x-y plane. Before the collision there was no momentum.

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

Due to conservation of momentum equation is written as,

$\begin{aligned}100{v_x}+2\left({200}\right)&=0\\100{v_x}&=-400\\{v_x}&=-4{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{aligned}$

The another equation from conservation of momentum is written as,

$\begin{aligned}100{v_y}+1.5\left({235}\right)&=0\\100{v_y}&=-352.5\\{v_y}&=-3.525{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{aligned}$

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

$\begin{aligned}v&=\sqrt{v_x^2+v_y^2}\\&=\sqrt{{{\left({-4}\right)}^2}+{{\left({-3.525}\right)}^2}}\\&=5.33{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{aligned}$

Therefore, the speed of third piece is $5.33{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}$ .

Thus, the speed of third piece that weighs $100{\text{ g}}$  is $\boxed{5.33{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}$ .

Learn More:

1. Speed and momentum brainly.com/question/6955558

2. Linear momentum brainly.com/question/11947870

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

Answer Details:

Grade: High School

Subject: Physics

Chapter: Momentum

Keywords:

Plate, momentum, falls, collision, slide, impact, 200 g, 2 m/s, 235 g, 1.50 m/s, 100 g, direction, motion, horizontal, force, x-axis, y-axis, vertically, floor, breaks, three, pieces.