Awọn ijamba inelastic apakan ni iwọn kan - awọn iṣoro ati awọn solusan

1. A 500-gram object, A, moving at 10 m / s and 200-gram object, B, moving at 12 m/s. The objects approach each other and collide. If the speed of object A after the ijamba is 6 m/s, what is the speed of object B after the collision?

A mọ̀:

ibi of object 1 (m1) = 500 giramu = 0.5 kg

ibi ti ohun 2 (m2) = 200 giramu = 0.2 kg

Ni ibẹrẹ siko of object 1 (v1) = -10 m/s

Initial velocity of object 2 (v2) = 12 m/s

The final velocity of object 1 (v1') = 6 m/s

Àwọn àmì àfikún àti àyọkúrò fi hàn pé àwọn ohun náà ń lọ sí ọ̀nà òdìkejì.

fe : the final velocity of object 2 (v2))

Ojutu:

m1 v1 + m2 v2 = m1 v1' + m2 v2'

(0.5)(-10) + (0.2)(12) = (0.5)(6) + (0.2)(v2))

-5 + 2.4 = 3 + 0.2 v2'

-2.6 = 3 + 0.2 v2'

-2.6 – 3 = 0.2 v2'

-5.6 = 0.2 v2'

v2’ = -5.6 / 0.2

v2' = -28 m/s

The speed of object 2 after collision is 28 m/s.

The plus and minus sign indicates that the objects move in the opposite direction.

Wo tun  Onínọmbà Oníwọ̀n - Àwọn ìṣòro àti àwọn ìdáhùn

2. Two equal-mass objects approach each other and collide. The speed of object 1 is 6 m/s and the speed of object 2 is 8 m/s. After the collision, object 2 moves leftward with speed of 5 m/s. What is the magnitude and direction of the velocity of object 1 after the ijamba?

Mọ:

ibi ti ohun 1 (m1) = m

ibi of object 2 (m2) = m

Initial speed of object 1 (v1) = -6 m/s

initial speed of object 2 (v2) = 8 m/s

The final speed of object 2 (v2') = -5 m/s

fe : the magnitude and direction of object 1 after collision

Ojutu:

Fọ́mùlà ti conservation of linear momentum :

m1 v1 + m2 v2 = m1 v1' + m2 v2'

mv1 + m v2 = m v1' + m v2'

m (v)1 +v2) = m (v1'+ v2))

v1 +v2 = v1'+ v2'

-6 + 8 = v1’ – 5

2 = v1’ – 5

2 + 5 = v1'

v1' = 7 m/s

The speed of object 1 after collision (v1)) is 3 m/s.

The plus and minus sign indicates that the objects move in the opposite direction.

Wo tun  Iṣipopada iyipo - awọn iṣoro ati awọn solusan

3. A 1-kg ball 1 and 2-kg ball 2 have the same direction and collide inelastically. Before the collision, ball 1 moves with speed of 10 m/s and ball 2 moves with speed of 5 m/s. The speed of ball 2 after the collision is 4 m/s. Determine the speed of ball 1 after the collision.

A mọ̀:

ibi rogodo 1 (m1) = 1 kg

ibi rogodo 2 (m2) = 2 kg

The speed of ball 1 (in1) = 10 m/s

The speed of ball 2 (v2) = 5 m/s

The final speed of ball 2 (v2') = 4 m/s

The plus sign of the velocity indicates that the balls have the same direction.

fe : the final velocity of ball 1 (v1))

Ojutu:

m1 v1 + m2 v2 = m1 v1' + m2 v2'

(1)(10) + (2)(5) = (1)(v1’) + (2)(4)

10 + 10 = v1' + 8

20 – 8 = v1'

v1' = 12 m/s

The speed of ball 1 after collision is 12 m/s.

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  1. Awọn iṣoro ati awọn solusan ti o wa ni ila
  2. Àwọn ìṣòro àti ìdáhùn ìṣíṣẹ́ àti ìṣíṣẹ́
  3. Awọn ijamba rirọ pipe ni awọn iṣoro ati awọn solusan iwọn kan
  4. Awọn ijamba inelastic pipe ni awọn iṣoro ati awọn solusan iwọn kan
  5. Awọn iṣoro ati awọn solusan ti ko ni ibatan si awọn ijamba ni iwọn kan

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