A-LevelPhysicsKinematics9702/22 · May/June 2017 · Q213 Marks
Define velocity.
- (a)Define velocity.[1]
- (b)(i)the horizontal velocity v as it leaves the table,[3]
- (b)(ii)the velocity just as it hits the floor,[4]
- (b)(iii)the kinetic energy just as it hits the floor,[2]
- (b)(iv)the loss in gravitational potential energy as it falls from the table to the floor.[2]
- (c)Explain why the kinetic energy of the ball in (b)(iii) does not equal the loss of gravitational potential energy in (b)(iv).[1]
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Mark scheme
Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.
- [B1] rate of change of displacement or change in displacement/time taken
- [C1] s = ut + ½at²
- [C1] t = [(2 x 1.25)/9.81]¹/² (= 0.5048 s)
- [A1] v = d / t = 1.5/0.50(48) = 3.0 (2.97) ms⁻¹
- [C1] vertical velocity = at = 9.81 × 0.5048 (= 4.95) [using t = 0.50 gives 4.9]
- [C1] velocity = [(vh)² + (vv)²]¹/²
- [A1] = [(2.97)² + (4.95)²]¹/² = 5.8 (5.79) [using t = 0.50 leads to 5.7]
- [A1] direction (= tan⁻¹ 4.95/2.97) = 59°
- [C1] kinetic energy = ½mv²
- [A1] = ½ × 0.45 × (5.8)² = 7.6 (7.57) J [using t = 0.50 leads to 7.3 J]
- [C1] potential energy = mgh
- [A1] = (0.45 x 9.81 × 1.25) = 5.5 (5.52) J
- [B1] there is KE of the ball at the start/leaving table
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