University Physics AI No. 7 Impulse, Momentum, and Collisions Class

Number

Name

I.Choose the Correct Answer 1. An object is moving in a circle at constant speed v. The magnitude of the rate of change of ( C

momentum of the object (A) is zero

(B) is proportional to v.

2



3

(C) is proportional to v .

(D) is proportional to v .

Solution: The magnitude of the centripetal acceleration is constant for uniform circular motion, that is

ac =

v2 = constant , r

v v dptotal , we have Using the general statement of Newton’s law of motion Ftotal = dt v dptotal v v2 = Ftotal = mac = m ∝ v 2 dt r 2. If the net force acting on a body is constant, what can we conclude about its momentum?

r

(A) The magnitude and/or the direction of p may change.

r (B) The magnitude of p remains fixed, but its direction may change. r (C) The direction of p remains fixed, but its magnitude may change. r (D) p remains fixed in both magnitude and direction.

( A )

Solution:

v

Using the general statement of Newton’s law of motion Ftotal =

v dptotal , we can know the dt

r

magnitude and/or the direction of p may change.

r

r

3. If I is the impulse of a particular force, what is dI / dt ?

( C

(A) The momentum (B) The change in momentum (C) The force (D) The change in the force Solution:

r r According the definition of the impulse, dI = Fdt ⇒

r r dI =F. dt



4. A variable force acts on an object from t i = 0 to t f . The impulse of the force is zero. One can conclude that

r r (A) ∆r = 0 and ∆p = 0 . r r (C) possibly ∆r ≠ 0 but ∆p = 0 .

r r (B) ∆r = 0 but possibly ∆p ≠ 0 .



r r (D) possibly ∆r ≠ 0 and possibly ∆p ≠ 0

Solution: According to the impulse-momentum theorem

( C



tf

ti

r r Fdt = ∆p ,the impulse of the force is zero, the

r change of the momentum ∆p = 0 . 5. A system of N particles is free from any external forces. Which of the following is true for the ( B )

magnitude of the total momentum of the system? (A) It must be zero. (B) It could be non-zero, but it must be constant. (C) It could be non-zero, and it might not be constant. (D) The answer depends on the nature of the internal forces in the system. Solution: According to the general statement of Newton’s law of motion

v Ftotal = r F = 0,

v dptotal , dt r ptotal = constant

II. Filling the Blanks 1. Fig.1 shows an approximate representation of force versus time during the collision of a

Fmax

58g tennis ball with a wall. This initial force (N)

velocity of the ball is 32m/s perpendicular to the wall; it rebounds with the same speed, also perpendicular to the wall. The value of Fmax , the maximum contact force during the collision is

928N . 0

Solution: Applying to the impulse-momentum theorem

1

2

Fig. 1

v v v I = ∆p = ∫ Fdt

So

v v ∆p = 2mv = ∫ Fdt = S F ⇒ 2 × 58 × 10− 3 × 32 = Fmax = 928 N

3

time (ms)

(2 + 6) × 10− 3 × Fmax 2

4

5

6

g. 2

r u

2. A stream of water impinges on a stationary “dished” turbine blade, as shown in Fig.2. The speed of the water is u, both before and after it strikes the curved surface of the blade, and the mass of water striking the blade per unit time is constant at the value µ. The force exerted by the water on the blade is 2uµ .

r −u Fig. 2

Solution:

v

v

v

Using the impulse-momentum theorem: I = ∆p = Fave ∆t So

∆mu − (−∆mu ) = 2∆mu = Fave ∆t

Thus force exerted by the water on the blade is Fave =

2∆mu = 2uµ ∆t

θ

3. A 320 g ball with a speed v of 6.22 m/s strikes a wall at angle θ of 33.0

o

and then rebounds with the same speed and angle (Fig.3). It is in contact with the wall for 10.4 ms. (a) The impulse was experienced by the wall is 2.17Ns . (b) The average force exerted by the ball on the wall is 2.09×102N .

θ

m

Solution:

Fig.3

v v v Using the impulse-momentum theorem: I = ∆p = Fave ∆t So the impulse was experienced by the wall is

I = ∆px = mv sin θ − (−mv sin θ ) = 2mv sin θ = 2 × 0.32 × 6.22 × sin 33o = 2.17 N ⋅ s The average force exerted by the ball on the wall is

Fave x =

2.17 I = = 2.09 × 102 N −3 ∆t 10.4 × 10

4. The muzzle speed of a bullet can be determined using a device called a ballistic pendulum, shown in Figure 4. A bullet of mass m moving at speed v encounters a large mass M hanging vertically as a pendulum at rest. The mass M absorbs the bullet. The hanging mass (now consisting of M + m) then swings to some height h above the initial position of the pendulum as shown. The initial speed v′ of the pendulum (with the embedded bullet) after impact is

r m v

M +m mv . The muzzle speed v of the bullet is 2 gh . If h m m+M = 10.0 cm, M = 2.50 kg, and m = 10.0 g. The muzzle speed v of the bullet is Solution:

h M Fig.4 352m/s .

(a) Use conservation of momentum mv = ( m + M )v' ⇒ v' =

mv m+M

(b) Assume the position which the pendulum is at rest is zero potential energy position, use CWE theory.

M +m M +m 1 ( M + m)v' 2 = ( M + m) gh ⇒ v' = 2 gh ⇒ v = v' = 2 gh m m 2 (c) v =

M +m 10 × 10 −3 + 2.5 × 2 × 9.81 × 10 = 352 m/s 2 gh = m 10 × 10 −3

5. Two objects, A and B, collide. A has mass 2.0 kg, and B has mass 3.0 kg. The velocities before

r

r

the collision are viA = (15m/s)iˆ + (30m/s) ˆj and viB = ( −10m/s)iˆ + (5.0m/s) ˆj . After the

r

collision, v fA = ( −6.0m/s)iˆ + (30m/s) ˆj . The final velocity of B is 4iˆ + 5 ˆj ( m/s) . Solution: Applying the conservation of the momentum, we have

v v v v m A v Ai + m B v Bi = m A v Af + m B v Bf

So the final velocity of B is

1 m v v v v v v v (mAv Ai + mB vBi − mAv Af ) = A (v Ai − v Af ) + vBi vBf = mB mB 2 v ⇒ vBf = × (15iˆ + 30 ˆj + 6iˆ − 30 ˆj ) − 10iˆ + 5 ˆj = 4iˆ + 5 ˆj (m/s) 3

III. Give the Solutions of the Following Problems 1. A massive anchor chain of length l is held vertical from its top link so that its lowest link is barely in contact with the horizontal ground. The chain is dropped. When the top link has fallen a distance y (see Figure 5), show that the magnitude of the normal force of the ground on the chain is 3 mgy/l. Solution: Choose the anchor chain on the ground as the study object, which have length y and mass

m' =

Origin y



l

Fig.5

m y . The chain on the ground get the gravitation m' g ,the force N acted by the floor and l

impulse force T acted by the falling chain. So

m m yg + T − N = 0 ⇒ N = yg + T l l Choose dm as the study object: dm =

m dy l

When it reaches the ground its speed is: v =

2 gy

(dmg − T ' )dt = −

Using impulse-momentum theorem:

m dy 2 gy l

Since dmg