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Question 23M.2.SL.TZ1.1

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Date May 2023 Marks available [Maximum mark: 14] Reference code 23M.2.SL.TZ1.1
Level SL Paper 2 Time zone TZ1
Command term Calculate, Determine, Draw, Estimate, Show Question number 1 Adapted from N/A
1.
[Maximum mark: 14]
23M.2.SL.TZ1.1

A ball of mass 0.800 kg is attached to a string. The distance to the centre of the mass of the ball from the point of support is 95.0 cm. The ball is released from rest when the string is horizontal. When the string becomes vertical the ball collides with a block of mass 2.40 kg that is at rest on a horizontal surface.

(a)

Just before the collision of the ball with the block,

(a.i)

draw a free-body diagram for the ball.

[2]

Markscheme

Tension upwards, weight downwards ✓
Tension is clearly longer than weight ✓

 

Look for:

(a.ii)

show that the speed of the ball is about 4.3 m s−1.

[1]

Markscheme

v = 2×9.81×0.95  OR  = 4.32 «m s−1» ✓

 

Must see either full substitution or answer to at least 3 s.f.

(a.iii)

determine the tension in the string.

[2]

Markscheme

T − mg = Fnet  OR  T − mg = mv2r ✓

T «= 0.800 × 9.81 + 0.800×4.31720.95» = 23.5 «N» ✓

(b)

After the collision, the ball rebounds and the block moves with speed 2.16 m s−1.

(b.i)

Show that the collision is elastic.

[4]

Markscheme

Use of conservation of momentum. ✓
Rebound speed = 2.16 «m s−1» ✓
Calculation of initial KE = «12 × 0.800 × 4.3172» = 7.46 « J » ✓
Calculation of final KE = «12 × 0.800  × 2.162 + 12 × 2.40 × 2.162» = 7.46 «J» ✓
«hence elastic»

(b.ii)

Calculate the maximum height risen by the centre of the ball.

[2]

Markscheme

ALTERNATIVE 1
Rebound speed is halved so energy less by a factor of 4 ✓
Hence height is 954 =23.8 «cm» ✓

ALTERNATIVE 2
Use of conservation of energy / 12 × 0.800 × 2.162 = 0.800 × 9.8 × h
OR

Use of proper kinematics equation (e.g. 0 = 2.162 − 2 × 9.8 × h) ✓

h = 23.8 «cm» ✓


Allow ECF from b(i)

(c)

The coefficient of dynamic friction between the block and the rough surface is 0.400. 

Estimate the distance travelled by the block on the rough surface until it stops.

[3]

Markscheme

ALTERNATIVE 1
Frictional force is f«= 0.400 × 2.40 × 9.81» = 9.42 «N» ✓
9.42 × d = 12 × 2.40 × 2.162  OR  d = 5.59879.42✓
d = 0.594 «m» ✓

 

ALTERNATIVE 2
a = «fm = µg = 0.4 × 9.81 =» 3.924 «m s−2» ✓

Proper use of kinematics equation(s) to determine ✓

d = 0.594 «m» ✓

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