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MakcuM [25]
3 years ago
6

Mitch throws a 100-g lump of clay at a 500-g target, which is at rest on a horizontal surface. After impact, the target, includi

ng the attached clay, slides 2.1 m before stopping. If the coefficient of friction is μ = 0.50, find the speed of the clay before impact.
Physics
1 answer:
max2010maxim [7]3 years ago
3 0

Answer:

27.22 m/s

Explanation:

Let the speed of clay before impact is u.

the speed of clay and target is v after impact.

use conservation of momentum

momentum before impact  momentum after impact

mass of clay x u = (mass of clay + mass of target) x v

100 x u = (100 + 500) x v

u = 6 v .....(1)

distance, s = 2.1 m

μ = 0.5

final velocity is zero. use third equation of motion

v'² = v² + 2as

0 = v² - 2 x μ x g x s

v² = 2 x 0.5 x 9.8 x 2.1 = 20.58

v = 4.54 m/s

so by equation (1)

u = 6 x 4.54 = 27.22 m/s

thus, the speed of clay before impact is 27.22 m/s.

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Answer:

1. G.P.E = 24 J

2. center of mass

Explanation:

Given the following data;

Mass = 2kg

Height, h = 1.2m

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To find the gravitational potential energy;

Gravitational potential energy (GPE) is an energy possessed by an object or body due to its position above the earth.

Mathematically, gravitational potential energy is given by the formula;

G.P.E = mgh

Where;

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  • h represents the height measured in meters.

Substituting into the formula, we have;

G.P.E = 2*1.2*9.8

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Translation kinetic energy is defined as the energy of a system due to the motion of the system’s center of mass. The center of mass is typically where the mass of the object or particle is concentrated.

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3 years ago
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F = 5.93×10^{13}\:\text{N}

Explanation:

Given:

m_1= 2×10^{16}\:\text{kg}

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Using Newton's universal law of gravitation, we can write

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\:\:\:\:= 5.93×10^{13}\:\text{N}

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