HINTS FOR PROBLEM SET #10

Problem Set #10 – Problem 1

This is a system with two motions and one degree of freedom. 

You can get the relationship between

the velocity of A and the velocity of B

by considering the length of string as a constant.

              2v_A = -v_B

(This equation assumes that x_A and x_B are both positive upward. )

(If you are having trouble deriving this relationship,

refer to Problem Set #2 at the bottom of the page:

 "The Conservation of String"  Explained)

This problem asks you to find a time interval

and you don’t care about finding acceleration.

Hence use the Principle of Impulse and Momentum.

              L₁ + I_t1→t2  =  L₂

Note:

              - this is a vector equation

              - t₁ and t₂  refer to instants of time

              - L = mv

              - I_t1→t2  is the sum of all the impulses acting between t₁ and t₂

              - If you assume a force, F, is constant over the time interval,

                   then its impulse will just be F∆t.

Let t₁ be the instant when the system is released (from rest). 

Let t₂ be the instant when the velocity of A reaches 2 ft/s. 

To calculate I_t1→t2 you need to know all the forces that act.

Draw a FBD for each body, A and B.

On A the forces acting include:

              - its weight, m_Ag

              - the normal force from the column, N

              - the force from the pulley, 2T

Apply the Principle of Impulse and Momentum to A in the vertical direction:

             0 + (- m_Ag∆t + 2T∆t) =  m_Av_A

Call this equation 1.

On B the forces acting include

              - its weight, m_Bg

              - the tension from the rope, T

Apply the Principle of Impulse and Momentum to B in the vertical direction:

             0 + ( - m_Bg∆t + T∆t) = - m_Bv_B

Call this equation 2.

Reduce the system of two equations (equation 1 and equation 2)

to a system of one equation that eliminates the tension, T.

Solve this equation for the time interval, ∆t.

You will also need the relationship you obtained between v_A and v_B at the beginning of this problem.
 

Problem Set #10 – Problem 2

Part a)

Use the Principle of Impulse and Momentum.

               L₁ + I_t1→t2  =  L₂

Let t₁ be the instant just before the 3 second interval.

Let t₂ be the instant just after the 3 second interval.

Draw a FBD for each person, mother and child. 

Both FBDs will include

              - their weights,  m_momg,  m_childg

              - the normal force from the ground, N_mom, N_child

              - the tension from the rope, T  (T_{mom =} - T_child = T )

Apply the Principle of Impulse and Momentum for each person, mother and child.

You will get two equations from which you can eliminate the tension in the rope T.

Then you can solve for the final speed of the mother.  

Part b)

To get the tension in the rope, just use the equation that you got by applying the Principle of Impulse and Momentum to the child’s motion.  Solve for T.

Problem Set #10 – Problem 3

Apply the Principle of Impulse and Momentum

to the ball in the horizontal direction. 

              L₁ + I_t1→t2  =  L₂

where

              - this is a scalar equation since you are looking at a specific direction

              - t₁ is the instant just before the ball hits the glove

              - t₂ is the instant the ball and glove come to a stop

                   - L₁ = m_ballv₁

              - L₂ = m_ballv₂ = 0

              - I- _t1→t2 = F_ave∆t

              - ∆t = distance F/ rate = (0.5 ft) / (30ft/s)  = 0.0167 sec

Plug these into the equation and solve for  F_ave.

Be careful of the units on both m_ball and v₁.

m_ball is not given, however,

you are given the ball's weight in ounces.

First convert ounces to pounds: 16oz = 1 lb_f

Then get the equivalent mass of that weight.

(Just divide by 32.2 ft/s².)

Also:

v₁  = 90 mph x (5280 ft/mile) x (1 hour/3600 sec) = 132 ft/s.