1989-12-27 — Page 24

9

証。

育日十三月一十年巳己曆麿夏

1990 中學會考預習專欄

明德出版社

MILL & DALE PRESS

Physics(13)

Exercise 13 Momentum

C.Y.Mak

(where necessary, take 9 = 10 2

1. As shown a figure 1(e),.a block of travelling with speed

MASS

Figure 2(b)

If the trolleys stuck together alter collision, find

(a) their commun velocity.

(b) the clergy lost in the impact.

3x105 3. A gun of mass

kg which can move freely on a horizontal ground, fires a grell of mass 103 kg with a nuzzle speed of 1800 ms If the Bystan (the gum and the shell) is at rest before firing, find

(a) the romentum of the system

before firing.

-

15 ms collides with another block af mass 6 ky traveling in oppusite

-1 direction with speed 5 ns

(b) the momentum of the system after

Firing.

If

after impact, the 4 kg-block returns

-1

(c) the recoil velocity of the gun

just after firing.

報日僑華

velocity!

A

三期星

The trolleys will move to e right with speed 8.4 ms"

[3

(b) Loss in K.E.

- ((^)(33) = g(6)(8)*]

:

日七廿月二十(九八九一年八十七國民華中 (24)

6. (a) Take the direction towards the

right as positive.

Let the velocity of A. after the

-1

collision be V ms

Initial momentum of the system

- $14.6) 18.4)*

- (12)(4) - (3)(-8)

= 2017.2 3

[Ans.)

= 24 kgs-1

after collision

Figure 4

(a) What is the velocity (magnitude and direction) of A after the collision?

(b). Calculate the

average (magnitude and direction) on A during the collision.

Torce

(b) By conservation

(c) Calculate the

force average Imagmtude and direction) on 3 Compare during the collision. the result with that of (b).

(d) Calculate the fraction of energy lost during the impact. In what form will the lost energy be dissipated2

with speed 3 ms. as shown in figure 1601, find

4. A ball of mass 2 kg is projected

horizontally

12 ms towards 咩 vortica! wal anc is rebounced back norizontally with

Solutions

1. (a) laking the direction

right as positive Momentum before impact

= (4)(15) (6) (-6)

服務好。

4 kg

2012

「機票

拿大

美

國

-1

= 24 kgs

6 kg

if the time of contact during the collision is 0.1 s.fini

= (4)[−3)+(61)

-6v-12

3. (a) Since the system is at rest Before firing, thus total momentum is zero.

of Linear momentum, total momentum after timng is also zero

(c) Let the velocity of the gun je v

after firing.

.. (3x107).

+ (102)(1800) = 0

-1

(Ans.)

The negative sign indicates the gun will recoil.

Final momentum of the system

= (12)(v) + (3)(4)

12v 12

Byteservation of momentury:

24 = 12v + 12

y = 1

the velocity of A is towards the right.

(h) The average force on A

as

rate of change of momentum of

(12)(1) - (12)(4)

-72N

0.5

Ans: The average force on A is

72N towards the left.

4. (a) Momentum

the ball

impact

(2)(12)

-1

= 24 gms

Momentum

of

Dall after

impact

(c) The average force on

Momentum after impact

- (2)(-10)

-1

= -20 kyus

Change in momentum

f

iinear

= (-20)-24

Immentum

24 - 6x-12

v = 6 MS

the

6ka-block will

move with

(The negative. sign ondicates ir. opposite direction).

Magnitude of Change in rementum is 44 xams.

(b) Magnitude of average force

Magnitude of Rate of change of aomentun

rate of change of momentum of

(3)(4) (3)(-8)

= 72N

0.5

the average force on B is 72N towards the right.

你

Compared with the result of, (b), thet the average force exerted on A by B is equal ragnitude with the average. force exerted on 3 by a but apposite in direction:

it Thus,

satisfies Newton's 3rd Law of Motion.

smooth horizontal floor

Figure Mal

9151

After Impact

Figure 1(b)

(a) the speed of the 6 kg block

after pact.

酒店

(b) calculate the erangy lost during

the impact.

{c}` what

form will energy be dissipated?

the lost

2. As shown in figure 2(a) and 2(b), the trolleys are travelling 你 smooth horizontal table before and after impact.

Belare Impact

(a) change in momentum of the ball before and after collision.

(h) the force. acting on the wall

churing coilision.

5. A ball of mass 0.4 kg is. released from 我 height. of 1.8 monito horizontal floor. It rebounds to a height of 0.8 m.

(a) Calculate the velocity of the ball just before and after impact-

(b) If the ball is in contacteswit

the floor for 0.02 s.

the magnitude of

the average

force exerted by the floor, o the space in vidct?

-1

6. Two trolleys A and move towards

and pach other with velocity dams 8.0 respectively. The mass of tro e kg and that of 8-15 3 kg. The two trolleys remain in contact for. 0.5 seconds during the head-on collision.

After collision, B rebounds with velocity of

figure 4.

as

5 as to the right.

(b) K.F. before collision

1.

=

任

K.Ffter collision

-441(31-45)(6)*

= 126 J

.. Loss in energy = 558 - 126

432 J (Ans.):

(c) Mainly converted into heat and

Sound energy.

the

2. (a) Taking

in

direction Lo

the

the right as positive Momentum before collision

= (4)(33)+(6)(-9)

84 kgrs

-1

Momentum after collision

063

(20

$ ras

-·

2.kq

A 3 Ky

2

(4-6) V

före collision

= 10 V

(Ans.)

ity before impact

6 ms

(Ans)

Velocity jasteafter impact

=

-1

= 4 ms (upwards)

[Ans.)

(b) Magnitude of Rate of change of

momentum

(0.41(4)

= 200N

(0.4)(6)

The average force required

= 200N + weight at the bat!

- 200+ (0291(10)

is

no

(d) Since there

change in potential energy before and after the impact, therefore the energy loss in this collision is ? 动 the kinetic energy loss.

Initial kinetic energy

-412/418+ (3:18)4

132 J

Final kinetic energy

=√(12) (112 - (3) (4)2

= 30 J

The fraction of energy lost

192 - 30

=

192

27

-32

(Ans.)

01186

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