Momentum and impulse Problems and Solutions 2

 Problem#1

A tennis player receives a shot with the ball (0.060 0 kg) traveling horizontally at 50.0 m/s and returns the shot with the ball traveling horizontally at 40.0 m/s in the opposite direction. (a) What is the impulse delivered to the ball by the racquet? (b) What work does the racquet do on the ball?

Answer:
Assume the initial direction of the ball in the –x direction.

Given: v_i = –50.0i m/s, v_f = 40.0i m/s, then

(a) the impulse delivered to the ball by the racquet is

I = ∆p = m(v_f – v_i)

I = (0.0600 kg)[(40.0i m/s) – (–50.0i m/s)] = 5.40i N.s

(b) work does the racquet do on the ball is

W = ∆K = K_f – K_i
W = ½(0.0600 kg)[(40.0 m/s)² – (–50.0 m/s)²] = –27.0 J

Problem#2
In a slow-pitch softball game, a 0.200-kg softball crosses the plate at 15.0 m/s at an angle of 45.0° below the horizontal. The batter hits the ball toward center field, giving it a velocity of 40.0 m/s at 30.0° above the horizontal. (a) Determine the impulse delivered to the ball. (b) If the force on
the ball increases linearly for 4.00 ms, holds constant for 20.0 ms, and then decreases to zero linearly in another 4.00 ms, what is the maximum force on the ball?

Answer:
Take x-axis toward the pitcher, then

(a) Impulse on the x-axis is given by

I_x = p_fx – p_ix

I_x = m[v_f cos30.0⁰ – v_i (–cos45.0⁰]

I_x = (0.200 kg)[(40.0 m/s) cos30.0⁰ + (15.0 m/s)(cos45.0⁰)]

I_x = 9.05i N.s

Impulse on the y-axis is given by

I_y = p_fy – p_iy

I_y = m[v_f sin30.0⁰ – v_i (–sin45.0⁰]

I_y = (0.200 kg)[(40.0 m/s) sin30.0⁰ + (15.0 m/s)(sin45.0⁰)]

I_y = 6.12j N.s

So, the impulse delivered to the ball is

I = (9.05i + 6.12j)N.s

(b) force graphs for time under these circumstances are shown below,

by using the graph above, the maximum force on the ball is

I = ½(0 + F_m)(4.00 x 10^-3 s) + F_m(20.0 x 10^-3 s) + ½ (F_m + 0)(4.00 x 10^-3 s)

(9.05i + 6.12j)N.s = (24.0 x 10^-3 s)F_m

F_m = (377i + 255j) N

Problem#3
A professional diver performs a dive from a platform 10 m above the water surface. Estimate the order of magnitude of the average impact force she experiences in her collision with the water. State the quantities you take as data and their values.

Answer:
If the diver starts from rest and drops vertically into the water, the velocity just before impact is
found from

K_f + U_f = K_i + U_i

½ mv² + 0 = 0 + mgy

v² = 2(9.80 m/s²)(10.0 m)

v = 14.0 m/s

With the diver at rest after an impact time of ∆t , the average force during impact is given by
F_avr∆t = m(v_f – v_i)

F_avr(1.00 s) = (55.0 kg)[0 – 14.0 m/s]

F_avr = 770 N (upward)

Problem#4
A garden hose is held as shown in Figure 2. The hose is originally full of motionless water. What additional force is necessary to hold the nozzle stationary after the water flow is turned on, if the discharge rate is 0.600 kg/s with a speed of 25.0 m/s?

Answer:
The force exerted on the water by the hose is

F_avr∆t = ∆p_water = m(v_f – v_i)

F_avr(0.10 s) = (0.600 kg)(25.0 m/s – 0)

F_avr = 15.0 N

According to Newton's third law, the water exerts a force of equal magnitude back on the hose.
Thus, the gardener must apply a 15.0 N force (in the direction of the velocity of the exiting water stream) to hold the hose stationary.

Problem#5
A glider of mass m is free to slide along a horizontal air track. It is pushed against a launcher at one end of the track. Model the launcher as a light spring of force constant k compressed by a distance x. The glider is released from rest. (a) Show that the glider attains a speed of v = x(k/m)^1/2. (b) Does a glider of large or of small mass attain a greater speed? (c) Show that the impulse imparted
to the glider is given by the expression x(km)^1/2. (d) Is a greater impulse injected into a large or a small mass? (e) Is more work done on a large or a small mass?

Answer:
(a) Energy is conserved for the spring-mass system:

K_f + U_f = K_i + U_i

½ mv² + 0 = 0 + ½ kx², then

v = x(k/m)^1/2

(b) From the equation, a smaller value of m makes v = x(k/m)^1/2 larger.

(c) I = ∆p = │p_f - p_i│ = mv

I = xm(k/m)^1/2

I = x(km)^1/2

(d) From the equation, a larger value of m makes I = x(km)^1/2 larger.

(e) For the glider, W = K_f – K_i = ½mv² – 0 = ½ kx²

The mass makes no difference to the work. 

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