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1 year ago

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Physics

2 answers:

Soloha48 [4]1 year ago

5 0

The momentum of the second ball was 15 kg.m/s.

<h3>What is inelastic collision?</h3>

In which collision some amount of kinetic energy of the system is lost that called inelastic collision. In purely inelastic collision, two bodies stick together. But principle of conservation of linear momentum is obeyed.

In the given question,

Two balls collide and after collision, the final momentum of the system = 18 kg.m/s.

Initial velocity of 1st ball of mass 3 kg is 1 m/s.

So, Initial momentum of first ball = mass × velocity = (3 kg) × (1 m/s) = 3 kg.m/s.

According to Principle of conservation of linear momentum for this inelastic collision,

Initial momentum of first ball + initial momentum of second ball = final momentum of the system

⇒ initial momentum of second ball = final momentum of the system - Initial momentum of first ball

= 18 kg.m/s - 3 kg.m/s.

= 15 kg.m/s.

Hence, initial momentum of second ball = 15 kg.m/s.

Learn more about momentum here:

brainly.com/question/24030570

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liq [111]1 year ago

5 0

Answer:

the answer is 15kgm/s

Ptotal=18 kgm/

m1v1+m2v1=18

(3kg)(1)+m2v1=18

3+m2v1=18

subtract 3

m2v1=15

because p=mv

the momentum of the second ball is 15 kgm/

Explanation:

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A playground slide is inclined 40°. If a boy with a mass of 32 kg slides down for -3meters. How much work is done by gravity on

telo118 [61]

Answer:

the work done by gravity on the boy is 604.62 J

Explanation:

Given;

distance the boy slides, d = 3 m

angle of inclination of the playground, θ = 40⁰

mass of the boy, m = 32 kg

The vertical height, h, above the ground through which the boy falls represents the height of the triangle which is the opposite side.

The distance through which the boy slides, d, represents the hypotenuse side of the right triangle.

$sin \theta = \frac{opposite }{hypotenuse} \\\\sin \theta = \frac{h}{d} \\\\h = dsin\theta\\\\h = 3 \times sin(40^0)\\\\h = 1.928 \ m$

The work done by gravity on the boy is calculated as;

W = P.E = mgh

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= 604.62 J

Therefore, the work done by gravity on the boy is 604.62 J

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A small object with momentum 7.0 kg∙m/s approaches head-on a large object at rest. The small object bounces straight back with a

EastWind [94]

Answer:

The magnitude of the large object's momentum change is 3 kilogram-meters per second.

Explanation:

Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:

$p_{S,1}+p_{B,1} = p_{S,2}+p_{B,2}$ (1)

Where:

$p_{S,1}$ , $p_{S,2}$ - Initial and final momemtums of the small object, measured in kilogram-meters per second.

$p_{B,1}$ , $p_{B,2}$ - Initial and final momentums of the big object, measured in kilogram-meters per second.

If we know that $p_{S,1} = 7\,\frac{kg\cdot m}{s}$ , $p_{B,1} = 0\,\frac{kg\cdot m}{s}$ and $p_{S, 2} = 4\,\frac{kg\cdot m}{s}$ , then the final momentum of the big object is: