Misalai 8 na Tambayoyin Momentum na Layi
1. Abu mai nauyin kilogiram 1 yana motsawa a gudun mita 10/s. lokacinta abin shine…
Tattaunawa
An sani :
Mass (m) = 1 kg
Gudun (v) = 10 m/s
An tambaya : ƙarfin layi (p)
Jawab :
Tsarin Momentum: p = mv
Bayani: p = ƙarfin motsi, m = taro, v = gudun.
Motsin abu shine:
p = mv = (1)(10) = 10 kg m/s2
2. Abubuwa biyu, kowannensu yana da nauyin kilogiram 2 da kilogiram 4, suna tafiya a gudun mita 20/s. Motsin kowane abu shine…
Tattaunawa
An sani :
Nauyin abu A (m)A) = 2 kg
Nauyin abu B (m)B) = 4 kg
Saurin abu A (v)A) = 20 m/s
Saurin abu B (v)B) = 20 m/s
An tambaya : ƙarfin abu A (pA) da kuma ƙarfin abu B (pB)
Jawab :
Motsin abu A :
pA = mA vA = (2)(20) = kilogiram 40 m/s
Motsin abu B :
pB = mB vB = (4)(20) = kilogiram 80 m/s
Idan saurin dukkan abubuwa guda biyu iri ɗaya ne, to abin da ke da babban nauyi zai sami ƙarin ƙarfin motsi.
3. Abubuwa A da B kowannensu yana da nauyin kilogiram 2. Abu A yana motsawa a gudun mita 2/s kuma abu B yana motsawa a gudun mita 4/s. Motsin abu A da abu B shine…
Tattaunawa
An sani :
Nauyin abu A (m)A) = 2 kg
Nauyin abu B (m)B) = 2 kg
Saurin abu A (v)A) = 2 m/s
Saurin abu B (v)B) = 4 m/s
An tambaya : ƙarfin abu A (pA) da kuma ƙarfin abu B (pB)
Jawab :
Motsin abu A :
pA = mA vA = (2)(2) = kilogiram 4 m/s
Motsin abu B :
pB = mB vB = (2)(4) = kilogiram 8 m/s
Idan tarin abubuwa biyu iri ɗaya ne, to abin da ke da saurin da ya fi girma zai sami ƙarfin motsi mafi girma.
4. Abu mai nauyin kilogiram 5 yana hutawa. Motsin abin shine…
Tattaunawa
An sani :
Nauyin abu (m) = 5 kg
Saurin abu (v) = 0
An tambaya : ƙarfin abin (p)
Jawab :
p = mv = (5)(0) = 0
Idan abu yana hutawa, to komai girman nauyin abu, ƙarfin abin ba shi da yawa.
5. A wasan ƙwallon baseball, ƙwallon da nauyinta ya kai kilogiram 0,5 tana motsawa da farko a gudun 2 ms -1 . Sannan ana buga ƙwallon da ƙarfi F a akasin motsin ƙwallon, don haka saurin ƙwallon ya canza zuwa 6 ms -1 . Idan ƙwallon ta taɓa jemage na tsawon daƙiƙa 0,01, to canjin ƙarfin shine...
A. 8 kg ms -1
B. 6 kg ms -1
C. 5 kg ms -1
D. 4 kg ms -1
E. 2 kg ms -1
Tattaunawa
An san cewa:
Nauyin ƙwallon (m) = 0,5 kg
Saurin farko (v o ) = 2 m/s
Saurin ƙarshe (v t ) = -6 m/s
Tazarar lokaci (t) = daƙiƙa 0,01
An tambaya: Canjin yanayin aiki
Amsa:
Tsarin canza yanayin motsi:
∆ p = mv t – mv o = m (v t – v o )
∆ p = (0,5 kg) (- 6 m/s - 2 m/s)
∆ p = (0,5 kg)(-8)
∆ p = 4 kg m/s
Amsar da ta dace ita ce D.
6. Abu mai nauyin gram 100 yana motsawa a gudun 5 ms -1 . Don tsayar da abu, ƙarfin juriya F yana aiki na 0,2 s. Girman ƙarfin F shine...
A. 0,5 N
B. 1,0 N
C. 2,5 N
D. 10 N
E. 25 N
Tattaunawa
An san cewa:
Nauyin abu (m) = gram 100 = 100/1000 = 0,1 kg
Saurin farko na abu (vo o ) = 5 m/s
Saurin ƙarshe na abu (v t ) = 0
Lokacin karo (Δt) = daƙiƙa 0,2
Tambaya: Girman ƙarfin F
Amsa:
Ka'idar motsin motsin rai (pulse-momentum theorem) ta bayyana cewa motsin rai (pulse) daidai yake da canjin motsin rai (momentum).
I = ΔP
F (Δt) = m (v t – v o )
F (0,2) = 0,1 (0 – 5)
F (0,2) = -0,5
F = -0,5 / 0,2
F = -2,5N
Girman ƙarfin shine Newtons 2,5. Alamar korau tana nuna alkiblar ƙarfin. Alamar korau tana nuna cewa ƙarfin yana akasin motsin abu.
Amsar da ta dace ita ce C.
7. Ana sauke ƙwallon tennis mai nauyin gram 100 zuwa ƙasa daga tsayin santimita 20 ba tare da saurin farko ba. Bayan buga ƙasa, ƙwallon tana tsalle da saurin 1 ms -1 da (g = 10 ms -2 ). Canjin motsin da ƙwallon ta fuskanta shine...
A. 0,1 Ns
B. 0,3 Ns
C. 0,5 Ns
D. 0,8 Ns
E. 0,9 Ns
Tattaunawa
An sani cewa :
Nauyin ƙwallon (m) = gram 100 = 0,1 kg
Tsawo (h) = 20 cm = mita 0,2
Saurin gudu saboda nauyi (g) = 10 m/s 2
Saurin ƙwallon nan da nan bayan ya buga ƙasa (v t ) = 1 m/s
An tambaya : canjin motsin ƙwallon
Amsa :
Saurin ƙwallon kafin buguwa (v)o)
A ƙididdige saurin ƙwallon kafin a yi tasiri ta amfani da dabarar motsi ta faɗuwa kyauta. Idan aka yi la'akari da tsayin ƙwallon faɗuwa kyauta (h) = mita 0,2, ana buƙatar saurin gudu saboda nauyi (g) = 10 m/s 2 kuma ana buƙatar saurin ƙwallon lokacin da ya bugi ƙasa, saboda haka ana amfani da dabarar v 2 = 2 gh .
v 2 = 2 (10)(0,2) = 4
v = √4 = -2 m/s
Ana ba shi alama mara kyau saboda alkiblar motsin ƙwallon kafin ya yi karo da ƙasa akasin alkiblar motsin ƙwallon bayan ya yi karo da ƙasa.
Canji a cikin motsin ƙwallon (Δp)
Δp = mv t - mv o = m (v t - v o )
Δp = (0,1)(1 – (-2)) = (0,1)(1 + 2) = (0,1)(3) = 0,3 Daƙiƙa na Newton
Amsar da ta dace ita ce B.
8. Wani abu da ya fara hutawa ya fashe zuwa sassa biyu tare da rabon 3:2. An jefa ɓangaren da ya fi girma a gudun 20 ms -1 . Don haka saurin da aka jefa ƙaramin ɓangaren shine...
A. 13,3 ms –1
B. 20 ms –1
C. 30 ms –1
D. 40 ms –1
E. 60 ms –1
Tattaunawa
An san cewa:
Nauyin abu 1 kafin karo = m
Gudun abu 1 kafin karo = 0 (abu a hutawa)
Nauyin kashi na 1 bayan karo (m 1 ) = 3m
Nauyin kashi na 2 bayan karo (m 2 ) = 2m
Saurin sashi na 1 bayan karo (v 1 ') = 20 m/s
Tambaya: Saurin kashi na 2 bayan karo (v 2 ')
Amsa:
Tsarin dokar kiyaye ƙarfin motsi:
m 1 v 1 = m 1 v 1 '+ m 2 v 2 '
(m) (0) = (3m) (20) + (2m) v 2 '
0 = 60m + (2m) v 2 '
60m = -2m v 2 '
60 = -2 v 2 '
v 2 ' = -60/2
v 2 ' = -30 m/s
Alamar mara kyau tana nuna cewa alkiblar motsi na kashi na 2 ya saba da alkiblar sashi na 1. Misali, idan aka jefa kashi na 1 zuwa gabas, to kashi na 2 za a jefa shi zuwa yamma.
Amsar da ta dace ita ce C.
Tushen tambaya:
Tambayoyin Nazarin Fizik na Ƙasa ga Makarantar Sakandare ta Babbar Sakandare/Makarantar Sakandare ta Sana'a
