he speed of the toy boat after $8.0 \\text{ s}$.","markscheme":"- Calculates the acceleration of the boat using Newton's second law: $a = \\frac{F}{m} = \\frac{3.0}{2.0} = 1.5 \\text{ m/s}^2$ (or calculates impulse: $I = F \\times t = 3.0 \\times 8.0 = 24 \\text{ N s}$) M1\n- Uses a SUVAT equation to find final speed: $v = u + at = 0 + 1.5 \\times 8.0$ (or equates impulse to change in momentum: $24 = 2.0 \\times v$) M1\n- Obtains $v = 12 \\text{ m/s}$ A1","marks":3,"order":0,"markingRubric":null,"explanation":null,"options":[]}],"options":[],"explanation":null,"firstAttemptFullMarks":null},{"id":"637a2e5b-0088-41a7-8023-8b8722f516a7","revisionId":"51ecf389-d27f-4455-9d52-352fbacabfc8","questionNumber":null,"questionType":"LA","specification":"![A plastic crate resting on a flat, rough concrete floor. A horizontal rope is attached to the right side of the crate, extending horizontally to the right. Labels with leader lines point to the 'plastic crate', 'rough concrete floor', and 'horizontal rope'.](https://assets.mathsgenie.co.uk/aqa-qb/0269a88d-69fd-416d-a9e2-c46345b846e6.png)\n\nA heavy plastic crate is pulled across a rough concrete floor at a constant horizontal velocity by a horizontal rope.","difficulty":7,"difficultyBand":"medium","totalMarks":6,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"52bb743f-fa55-4838-9997-44d238f2dfe3","label":null,"content":"Explain how the forces acting on the plastic crate keep it moving at a constant horizontal velocity.\n\nYour answer should refer to all the vertical and horizontal forces acting on the crate.","markscheme":"- Identifies **weight** acting vertically downwards on the crate B1\n- Identifies **normal contact force** (or reaction force) acting vertically upwards from the floor B1\n- Identifies **pulling force** (tension) acting horizontally forwards and **frictional force** acting horizontally backwards B1\n- Explains that the vertical forces are equal and opposite (vertical resultant force is zero) as there is no vertical acceleration M1\n- States that a constant horizontal velocity implies zero horizontal acceleration ($a = 0$), so the horizontal resultant force is zero (by Newton's First Law) M1\n- Concludes that the forward pulling force is equal in magnitude and opposite in direction to the backward frictional force, giving a net resultant force of zero A1","marks":6,"order":0,"markingRubric":null,"explanation":{"methods":[{"id":"balanced_forces","label":"Balance the forces","steps":[{"type":"drawing","order":0,"title":"Draw the force diagram","width":825,"height":356,"shapes":[{"id":"crate","type":"line","color":"black","points":[{"x":213.3,"y":246.5},{"x":549.3,"y":246.5},{"x":549.3,"y":78.5},{"x":213.3,"y":78.5},{"x":213.3,"y":246.5}]},{"id":"normal","toX":325.3,"toY":246.5,"type":"arrow","color":"red","fromX":325.3,"fromY":313.7},{"id":"weight","toX":437.3,"toY":313.7,"type":"arrow","color":"red","fromX":437.3,"fromY":246.5},{"id":"friction","toX":101.3,"toY":162.5,"type":"arrow","color":"blue","fromX":213.3,"fromY":162.5},{"id":"tension","toX":661.3,"toY":162.5,"type":"arrow","color":"blue","fromX":549.3,"fromY":162.5},{"x":42,"y":263.6,"id":"normal-label","text":"normal contact force","type":"text","color":"red"},{"x":446.3,"y":263.6,"id":"weight-label","text":"weight","type":"text","color":"red"},{"x":110.1,"y":171.5,"id":"friction-label","text":"friction","type":"text","color":"blue"},{"x":566,"y":124.5,"id":"tension-label","size":"s","text":"pulling force (tension)","type":"text","color":"blue"},{"x":494.1,"y":42,"id":"label-1","text":"plastic crate","type":"text"}],"caption":"A clean free-body diagram of the crate. The opposite force arrows have equal lengths to show that the forces balance.","diagramSpec":{"points":[{"at":[0,0],"name":"A"},{"at":[6,0],"name":"B"},{"at":[6,3],"name":"C"},{"at":[0,3],"name":"D"},{"at":[2,-1.2],"name":"Nstart"},{"at":[2,0],"name":"Nend"},{"at":[4,0],"name":"Wstart"},{"at":[4,-1.2],"n

Newton's laws and momentum

Question 12

Newton's laws and momentum Questions

Newton's laws and momentum

Question 12

Newton's laws and momentum Questions