Question

A glass soda bottle is emptied of soda and filled to the very top with water. A cork is carefully fitted into the top of the bottle, leaving no air between the corkand the water. The top of the bottle has a diameter of D_top = 2.00 cm and the bottom of the bottle has a diameter of D_bot = 6.50 cm. The glass breaks when it isexposed to p_max = 70.0 MPa of pressure.
A student hits the cork sharply with her fist and the bottom of the bottle breaks. The student's fist has a mass of m = 0.480 kg and moves downward at a speed of v_i =5.00 m/s. It collides elastically with the cork and rebounds with the same speed. The collision lasts for t = 1.20×10-4 s. In this problem, the positive direction isupward.

What is the force that the fist exerts on the top of the bottle?
Express your answer in newtons.

Answer 1

Concepts and reason

The concept used to solve this problem is momentum change.

Initially, the change in momentum can be calculated by substituting the value of the mass and the velocity. Later, the force exerted on the top of the bottle can be calculated by using the relation which relates the force, momentum change and the time.

Fundamentals

The expression for the momentum change is,

Here, is the change in momentum, is the student’s fist, and is the velocity.

Here, the velocity is negative as the force is direction along downward direction.

The expression of the force exerts on the top of the bottle is,

Here, is the force exerts on the top of the bottle, is the change in time.

The expression for the change in momentum is,

Substitute for, for.

The expression for the force exerts on the top of the bottle is,

Substitute for, for.

Ans:

The force exerts on the top of the bottle is.

Answer 2

Given Rt=0.0100m, Rb=0.0325m, m=0.480kg, vi=-vf=5.00m/s, tc=120µs
Find top and bottom areas:
At = p(Rt)² = 0.0001p m²
Ab = p(Rb)² = 0.00105625p m²

Although water is incompressible, cork &fist are not. Assume deceleration is uniform and half the time of collision. First find force at top:
Ft = m*dv/dt
= m(vi-0)/(tc/2)
= (0.48)(5)/((120*10^-6)/2)
= 40 kN (or 9000 lbf)

Now find force at bottom:
Fb = Ft*Ab/At
= (40000)(0.00105625p)/(0.0001p)
= 422.5 kN

Convert force to pressure to see if exceeds bottle rating:
Pb = Fb/Ab = 127.3 MPa (or 18000 psi)
127.3 MPa > p_max= 70 MPa

Answer 3

According to continiuity equation
A1 * v1 = A2 * v2
or v2 = (A1/A2) * v1
According to Bernoulli's equation
P1 + ?gh1 + (1/2)?v1^2 = P2 + ?gh2 +(1/2)?v2^2
or P2 = P1 + ?g * (h1 - h2) + (1/2)? * (v1^2 -v2^2)
h1 = h2
or P2 = P1 + (1/2)? * (v1^2 - v2^2) = P1 + (1/2)? *[(1 - (v2/v1)^2)] = P1 + (1/2)? * v1^2 * [(1 -(A1/A2)^2)]
or P2 = P1 + (1/2)? * v1^2 * [(1 - (d1/d2)^2)]
P1 = 70.0 ,? = 1000 kg/m^3,v1 = 5.00 m/s,d1 = 2.00 cm = 2.00 *10^-2 m and d2 = 6.5 cm = 6.5 * 10^-2 m

Answer 4

Given Rt=0.0100m, Rb=0.0325m, m=0.480kg, vi=-vf=5.00m/s, tc=120µs
Find top and bottom areas:
At = p(Rt)² = 0.0001p m²
Ab = p(Rb)² = 0.00105625p m²

Although water is incompressible, cork &fist are not. Assume deceleration is uniform and half the time of collision. First find force at top:
Ft = m*dv/dt
= m(vi-0)/(tc/2)
= (0.48)(5)/((120*10^-6)/2)
= 4x10^4N

Answer 5

Given Rt=0.0100m, Rb=0.0325m, m=0.480kg, vi=-vf=5.00m/s, tc=120µs
Find top and bottom areas:
At = p(Rt)² = 0.0001p m²
Ab = p(Rb)² = 0.00105625p m²

Although water is incompressible, cork &fist are not. Assume deceleration is uniform and half the time of collision. First find force at top:
Ft = m*dv/dt
= m(vi-0)/(tc/2)
= (0.48)(5)/((120*10^-6)/2)
= 40 kN (or 9000 lbf)

Now find force at bottom:
Fb = Ft*Ab/At
= (40000)(0.00105625p)/(0.0001p)
= 422.5 kN

Convert force to pressure to see if exceeds bottle rating:
Pb = Fb/Ab = 127.3 MPa (or 18000 psi)
127.3 MPa > p_max= 70 MPa

Answer 6

Given Rt=0.0100m, Rb=0.0325m, m=0.480kg, vi=-vf=5.00m/s, tc=120µs
Find top and bottom areas:
At = p(Rt)² = 0.0001p m²
Ab = p(Rb)² = 0.00105625p m²

Although water is incompressible, cork &fist are not. Assume deceleration is uniform and half the time of collision. First find force at top:
Ft = m*dv/dt
= m(vi-0)/(tc/2)
= (0.48)(5)/((120*10^-6)/2)
= 40 kN (or 9000 lbf)

Now find force at bottom:
Fb = Ft*Ab/At
= (40000)(0.00105625p)/(0.0001p)
= 422.5 kN

Convert force to pressure to see if exceeds bottle rating:
Pb = Fb/Ab = 127.3 MPa (or 18000 psi)
127.3 MPa > p_max= 70 MPa

Answer 7

Concepts used in the answer:- definition of force as rate of change of linear momentum has been used for this answer...***************************************************************************************************
This concludes the answers. If there is any mistake, let me know immediately and I will fix it....