ards","topicId":"917e5e97-5e0a-471a-805a-2154fe7d21fe","topicSlug":"kinematics-9","cards":[{"id":"522ac667-fab3-461c-a48b-2b765df71d8c","cardType":"basic","front":"After launch, what force acts on a projectile in the standard model?","back":"Its weight only.","points":1,"mcqOptions":null},{"id":"c8161623-108c-4728-86d4-f8d64ed2ff3d","cardType":"basic","front":"Why does a projectile's horizontal velocity remain constant?","back":"There is no horizontal acceleration.","points":2,"mcqOptions":null},{"id":"ae5fe86b-e30c-461e-a6ef-a585cc412a41","cardType":"basic","front":"What quantity links a projectile's horizontal and vertical motions?","back":"The same elapsed time $t$.","points":1,"mcqOptions":null},{"id":"b12531de-b078-4d8c-8a74-ea8b76b81d2b","cardType":"basic","front":"If upward is positive, what is the vertical acceleration of a projectile?","back":"$-g$, usually $-9.8\\text{ m s}^{-2}$.","points":1,"mcqOptions":null},{"id":"21c72d63-5db5-48ed-9fd9-7b25e7349c79","cardType":"cloze","front":"For launch speed $u$ at angle $\\theta$ above the horizontal, $u_x=$ $\\text{[...]}$ and $u_y=$ $u\\sin\\theta$.","back":"For launch speed $u$ at angle $\\theta$ above the horizontal, $u_x=$ $\\boxed{u\\cos\\theta}$ and $u_y=$ $u\\sin\\theta$.","points":null,"mcqOptions":null},{"id":"1bca075c-b0d2-435f-9ac9-370d8883aacf","cardType":"cloze","front":"For launch speed $u$ at angle $\\theta$ above the horizontal, $u_x=$ $u\\cos\\theta$ and $u_y=$ $\\text{[...]}$.","back":"For launch speed $u$ at angle $\\theta$ above the horizontal, $u_x=$ $u\\cos\\theta$ and $u_y=$ $\\boxed{u\\sin\\theta}$.","points":null,"mcqOptions":null},{"id":"49f68e8b-a946-499e-a80d-1a9fd2d798be","cardType":"basic","front":"If upward is positive, what sign does $y$ have below the launch point?","back":"Negative: $y<0$.","points":1,"mcqOptions":null},{"id":"ecdf3a45-ba25-44d4-af2f-5d41abb8e2a8","cardType":"cloze","front":"Horizontal motion is described by $v_x=$ $\\text{[...]}$ and $x=$ $u\\cos\\theta\\,t$.","back":"Horizontal motion is described by $v_x=$ $\\boxed{u\\cos\\theta}$ and $x=$ $u\\cos\\theta\\,t$.","points":null,"mcqOptions":null},{"id":"e6c7828a-e060-4609-b401-fd45e357fcbc","cardType":"cloze","front":"Horizontal motion is described by $v_x=$ $u\\cos\\theta$ and $x=$ $\\text{[...]}$.","back":"Horizontal motion is described by $v_x=$ $u\\cos\\theta$ and $x=$ $\\boxed{u\\cos\\theta\\,t}$.","points":null,"mcqOptions":null},{"id":"d52eca9e-f91e-48fa-ae8c-aad37d3c4e44","cardType":"cloze","front":"With upward positive, vertical motion is described by $v_y=$ $\\text{[...]}$ and $y=$ $u\\sin\\theta\\,t-\\frac{1}{2}gt^2$.","back":"With upward positive, vertical motion is described by $v_y=$ $\\boxed{u\\sin\\theta-gt}$ and $y=$ $u\\sin\\theta\\,t-\\frac{1}{2}gt^2$.","points":null,"mcqOptions":null},{"id":"d430ad1d-5804-4d6a-baa1-05cf74383a90","cardType":"cloze","front":"With upward positive, vertical motion is described by $v_y=$ $u\\sin\\theta-gt$ and $y=$ $\\text{[...]}$.","back":"With upward positive, vertical motion is described by $v_y=$ $u\\sin\\theta-gt$ and $y=$ $\\boxed{u\\sin\\theta\\,t-\\frac{1}{2}gt^2}$.","points":null,"mcqOptions":null},{"id":"731e0ad4-2407-486b-a6b5-1d324b62968d","cardType":"basic","front":"What is zero at a projectile's highest point?","back":"Its vertical velocity: $v_y=0$.","points":1,"mcqOptions":null},{"id":"c83492d6-3fff-49b1-9791-9952e59b2ded","cardType":"basic","front":"What is the projectile's acceleration at its highest point, with upward positive?","back":"It remains $-g$.","points":1,"mcqOptions":null},{"id":"d3b9d777-9784-4b93-a0a3-eccbec872558","cardType":"basic","front":"What is the horizontal velocity at the highest point?","back":"It remains $u\\cos\\theta$.","points":1,"mcqOptions":null},{"id":"e00ac541-5bd8-488e-87f7-7cc904130baa","cardType":"basic","front":"For a projectile returning to its launch level, what is its time of flight?","back":"$T=\\frac{2u\\sin\\theta}{g

4.1.3 Kinematics (A-level only) Flashcards

4.1.3 Kinematics (A-level only) Flashcards