Respuesta :
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
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
The speed of third piece that weighs [tex]100{\text{ g}}[/tex] is [tex]\boxed{5.33{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}[/tex] .
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 [tex]200{\text{ g}}[/tex] moves along the x-axis by the speed of [tex]2{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex] and the other piece of [tex]235{\text{ g}}[/tex] moves on the y-axis by the speed of [tex]1.5{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex] .
Our aim is to find the speed of third piece that weighs [tex]100{\text{ g}}[/tex] .
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 [tex]{v_x}[/tex] and [tex]{v_y}[/tex] , and the resultant of velocity [tex]v[/tex] makes an angle of [tex]\phi[/tex] under the negative x-axis as shown in Figure 1.
Due to conservation of momentum equation is written as,
[tex]\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}[/tex]
The another equation from conservation of momentum is written as,
[tex]\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}[/tex]
The velocity [tex]v[/tex] by which the third piece moves is calculated as,
[tex]\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}[/tex]
Therefore, the speed of third piece is [tex]5.33{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex] .
Thus, the speed of third piece that weighs [tex]100{\text{ g}}[/tex] is [tex]\boxed{5.33{\text{ }}{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}[/tex] .
Learn More:
1. Speed and momentum https://brainly.com/question/6955558
2. Linear momentum https://brainly.com/question/11947870
3. Velocity and Momentum https://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.
