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lawyer [7]
3 years ago
14

A rubber ball thrown at a speed of 5 m/s hit a flat wall and

Physics
1 answer:
velikii [3]3 years ago
5 0

Answer:

See the explanion below

Explanation:

Linear momentum is defined as the product of mass by velocity and can be calculated by the following equation.

P =m*v

where:

P = momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

When the ball hits the wall it receives a force for a very short time, this time and force value is known as momentum and can be calculated by the following expression.

-(m_{1}*v_{1}) + (F*t)=(m_{1}*v_{2})

where:

m₁ = mass of the ball [kg]

v₁ = 5 [m/s]

v₂ = 5 [m/s]

v₁ = v₂

F*t = Imp

Imp = (m_{1}*v_{1})+ (m_{1}*v_{2})\\Imp = 2*m*v_{1}

It can be said that the impulse is equal to 2 times the speed of throwing by the mass.

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One object (m1 = 0.220 kg) is moving to the right with a speed of 2.10 m/s when it is struck from behind by another object (m2 =
blagie [28]

Answer:

vf₁  = 6.86 m/s , to the right

vf₂ =  2.96 m/s, to the right

Explanation:

Theory of collisions  

Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:  

p=m*v  

where  

p:Linear momentum  

m: mass  

v:velocity  

There are 3 cases of collisions : elastic, inelastic and plastic.  

For the three cases the total linear momentum quantity is conserved:  

P₀ = Pf Formula (1)  

P₀ :Initial linear momentum quantity  

Pf : Final linear momentum quantity  

Data

m₁= 0.220 kg : mass of  object₁

m₂= 0.345 kg : mass of  object₂

v₀₁ =  2.1 m/s ₁ , to the right : initial velocity of m₁

v₀₂=   6 m/s, to the right  i :initial velocity of m₂

Problem development

We appy the formula (1):

P₀ = Pf  

m₁*v₀₁ + m₂*v₀₂ = m₁*vf₁ + m₂*vf₂  

We assume that the two objects move to the right at the end of the collision, so, the sign of the final speeds is positive:

(0.22)*(2.1) + (0.345)*(6) = (0.22)*vf₁ +(0.345)*vf₂

2.532 = (0.22)*vf₁ +(0.345)*vf₂ Equation (1)

Because the shock is elastic, the coefficient of elastic restitution (e) is equal to 1.

e= \frac{v_{f2}-v_{f1} }{v_{o1} -v_{o2} }

1*(v₀₁ - v₀₂ )  = (vf₂ -vf₁)

(2.1 - 6 )  = (vf₂ -vf₁)

-3.9 =  (vf₂ -vf₁)

vf₂ = vf₁ - 3.9

vf₂ = vf₁ - 3.9 Equation (2)

We replace Equation (2) in the Equation (1)

2.532 = (0.22)*vf₁ +(0.345)*( vf₁ - 3.9)

2.532 = (0.22)*vf₁ +(0.345)* (vf₁) -(0.345)( 3.9)

2.532 + 1.3455 = (0.565)*vf₁

3.8775 = (0.565)*vf₁

vf₁  = (3.8775) / (0.565)

vf₁  = 6.86 m/s, to the right

We replace vf₁  = 6.86 m/s in the Equation (2)

vf₂ =  6.86 - 3.9

vf₂ =  2.96 m/s, to the right

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Note that we added 30 and 6 because it was going downstream. If it was going upstream, then we would have had to subtract.
8 0
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