Question 19M.2.SL.TZ2.1bii
| Date | May 2019 | Marks available | [Maximum mark: 3] | Reference code | 19M.2.SL.TZ2.1bii |
| Level | SL | Paper | 2 | Time zone | TZ2 |
| Command term | Show that | Question number | bii | Adapted from | N/A |
A student strikes a tennis ball that is initially at rest so that it leaves the racquet at a speed of 64 m s–1. The ball has a mass of 0.058 kg and the contact between the ball and the racquet lasts for 25 ms.
The student strikes the tennis ball at point P. The tennis ball is initially directed at an angle of 7.00° to the horizontal.
The following data are available.
Height of P = 2.80 m
Distance of student from net = 11.9 m
Height of net = 0.910 m
Initial speed of tennis ball = 64 m s-1
Show that the tennis ball passes over the net.
[3]
ALTERNATIVE 1
uy = 64 sin7/7.80 «ms−1»✔
decrease in height = 7.80 × 0.187 + × 9.81 × 0.1872/1.63 «m» ✔
final height = «2.80 − 1.63» = 1.1/1.2 «m» ✔
«higher than net so goes over»
ALTERNATIVE 2
vertical distance to fall to net «= 2.80 − 0.91» = 1.89 «m»✔
time to fall this distance found using «=1.89 = 7.8t + × 9.81 ×t2»
t = 0.21 «s»✔
0.21 «s» > 0.187 «s» ✔
«reaches the net before it has fallen far enough so goes over»
Other alternatives are possible
There were a number of approaches students could take to answer this and examiners saw examples of them all. One approach taken was to calculate the time taken to fall the distance to the top of the net and to compare this with the time calculated in bi) for the ball to reach the net. This approach, which is shown in the mark scheme, required solving a quadratic in t which is beyond the mathematical requirements of the syllabus. This mathematical technique was only required if using this approach and not required if, for example, calculating heights.
A common mistake was to forget that the ball has a vertical acceleration. Examiners were able to award credit/ECF for correct parts of an otherwise flawed method.
