Misalai 4 na Tambayoyin Sha'awa
1. Ƙwallo mai nauyin kilogiram 0,5 ta faɗi cikin sauƙi daga tsayin h 1 = mita 7,2 sama da ƙasa ta kuma yi tsalle zuwa tsayin h 2 = mita 3,2. Idan saurin gudu saboda nauyi ya kai ms 10 -2 , bugun da ke kan ƙwallon zai...
A. 2,0 Ns
B. 3,0 Ns
C. 10 Ns
D. 40 Ns
E. 80 Ns
Tattaunawa
An sani cewa :
Nauyin ƙwallon (m) = 0,5 kg
Tsayin ƙwallon da ke faɗuwa daga sama (h 1 ) = mita 7,2
Tsayin ƙwallon da ke tsalle (h 2 ) = mita 3,2
Saurin nauyi na duniya (g) = 10 m/s 2
An tambaya : Sha'awa tana aiki a kan ƙwallon
Amsa :
Saurin ƙwallon kafin ya yi karo (v)o)
Lissafin saurin ƙwallon kafin bugun ta amfani da dabarar faɗuwar 'free fall'. Idan aka yi la'akari da tsayin (h) = mita 7,2, saurin da nauyi ya haifar (g) = 10 m/s 2 da kuma saurin ƙarshe kafin ya taɓa ƙasa, yi amfani da dabarar v 2 = 2 gh
v o 2 = 2(10)(7,2) = 144
v o = 2(10)(7,2) = 12 m/s
Gudun ƙwallon kafin bugun (v o ) shine -12 m/s. Alamar korau tana nuna alkibla ne kawai.
Saurin ƙwallon nan da nan bayan karo (v)t)
Lissafin saurin ƙwallon nan da nan bayan karo ta amfani da dabarar motsi a tsaye sama. Idan aka yi la'akari da tsayin (h) = mita 3,2, saurin nauyi (g) = -10 m/s 2 , saurin ƙarshe a matsakaicin tsayi (v t 2 ) = 0 kuma ana buƙatar saurin farko nan da nan bayan buga ƙasa, don haka yi amfani da dabarar v t 2 = v o 2 + 2 gh
v t 2 = v o 2 + 2 gh
0 = v o 2 + 2 (-10)(3,2)
v o 2 = 64
v o = √64 = 8 m/s
Gudun ƙwallon bayan karo (v t ) shine 8 m/s
Sha'awar motsa jiki a kan ƙwallon (I)
Ana ƙididdige impulse ta amfani da dabarar theorem na impulse-momentum.
Impulse (I) = canjin motsi (Δp)
I = m (v t – v o ) = (0,5)(8-(-12)) = (0,5)(8 + 12) = 0,5(20) = 10 Newton daƙiƙa
Amsar da ta dace ita ce C.
2. Kwallon ping pong mai nauyin gram 5 ya faɗi cikin sauƙi daga wani tsayi (g = 10 ms -2 ). Idan ya faɗi ƙasa, saurin ƙwallon shine 6 ms -1 kuma nan da nan bayan ya faɗi ƙasa, ƙwallon tana tsalle sama da gudun 4 ms -1 . Girman bugun da ke kan ƙwallon shine...
A. 0,50 Ns
B. 0,25 Ns
C. 0,10 Ns
D. 0,05 Ns
E. 0,01 Ns
Tattaunawa
An sani cewa :
Nauyin ƙwallon (m) = gram 5 = 0,005 kg
Saurin ƙwallon kafin ya buga ƙasa (v o ) = -6 m/s
Saurin ƙwallon nan da nan bayan ya buga ƙasa (v t ) = 4 m/s
Gudu mai kyau da mara kyau yana nufin cewa alkiblar gudun ƙwallon ko alkiblar motsin ƙwallon kafin karo ya saba da alkiblar motsin ƙwallon bayan karo.
An tambaya : Impulse (I)
Amsa :
Ka'idar motsin motsin rai (pulse-momentum) ta bayyana cewa motsin rai (I) daidai yake da canjin motsin rai (Δp).
I = Δp = mv t - mv o = m (v t - v o )
I = (0,005)(4 – (-6)) = (0,005)(4 + 6) = (0,005)(10) = 0,05 Newton daƙiƙa
Amsar da ta dace ita ce D.
3. Ana jefa ƙwallon gram 20 da saurin v 1 = 4 ms -1 a hagu. Bayan buga bango, ƙwallon ta yi tsalle da saurin v 2 = 2 ms -1 a dama. Sakamakon bugun shine….
A. 0,24 Ns
B. 0,12 Ns
C. 0,08 Ns
D. 0,06 Ns
E. 0,04 Ns
Tattaunawa
An san cewa:
Nauyin ƙwallon (m) = gram 20 = 0,020 kg
Gudun ƙwallon kafin karo (v o ) = -4 m/s (zuwa hagu)
Saurin ƙwallon bayan karo (v t ) = +2 m/s (zuwa dama)
v o an ba shi alama mara kyau don bambanta alkiblarsa daga v t.
An tambaya: Impulse
Amsa:
Impulse (I) = canjin motsi (Δp) = mv t – mv o
Bugawa (I) = m (v t – v o ) = 0,02 (2 – (-4))
Sha'awar (I) = 0,02 (2 + 4) = 0,02 (6)
Impulse (I) = 0,12 Newton daƙiƙa.
Amsar da ta dace ita ce B.
4. Ana jefa 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 :
Nauyin ƙwallon (m) = gram 100 = 0,1 kg
Tsawo (h) = 20 cm = mita 0,2
Saurin gudu saboda nauyi (g) = 10 m/s2
Saurin ƙwallon nan da nan bayan ya buga ƙasa (v)t) = 1 m/s
An tambaya : canjin yanayin motsi na ƙwallon
Jawab :
Saurin ƙwallon kafin buguwa (v)o)
Lissafin saurin ƙwallon kafin tasiri ta amfani da dabarar faɗuwar 'free fall'. Idan aka yi la'akari da tsayin ƙwallon faɗuwar 'free ball' (h) = mita 0,2, saurin da aka samu sakamakon nauyi (g) = 10 m/s2 kuma ya nemi saurin ƙwallon lokacin da ya buga ƙasa, don haka aka yi amfani da dabarar v2 = 2 gh
v2 = 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 = mvt – mvo = m (v)t - vo)
Δp = (0,1)(1 – (-2)) = (0,1)(1 + 2) = (0,1)(3) = 0,3 Daƙiƙa na Newton
Amsar da ta dace ita ce B.
Tushen tambaya:
Tambayoyin Nazarin Fizik na Ƙasa ga Makarantar Sakandare ta Babbar Sakandare/Makarantar Sakandare ta Sana'a
