Standard +0.3 This is a standard M1 kinematics problem requiring students to set up SUVAT equations for two particles and use the condition that speeds are equal at collision. While it involves simultaneous equations and careful sign conventions, the approach is straightforward and well-practiced in M1 courses. The 'show that T=1.5' guidance significantly reduces the problem-solving demand.
Small stones A and B are initially in the positions shown in Fig. 6 with B a height \(H\) m directly above A.
\includegraphics{figure_5}
At the instant when B is released from rest, A is projected vertically upwards with a speed of \(29.4\text{ms}^{-1}\). Air resistance may be neglected.
The stones collide \(T\) seconds after they begin to move. At this instant they have the same speed, \(V\text{ms}^{-1}\), and A is still rising.
By considering when the speed of A upwards is the same as the speed of B downwards, or otherwise, show that \(T = 1.5\) and find the values of \(V\) and \(H\). [7]
If 2 subs there must be a statement about equality
FT T or V, whichever is found second
Sum of the distance travelled by each attempted
cao
Attempts at V2 for each particle equated. Allow
sign errors, 9.81 etc
Allow h , h without h =H−h
1 2 1 2
Both correct. Require h =H−h but not an
1 2
equation.
cao
Any method that leads to T or V
Any method leading to the other variable
Other approaches possible. If ‘clever’ ways seen,
Answer
Marks
reward according to weighting above.
7
7
Question 5:
5 | Mark | Comment | Sub
Method 1
↑ v =29.4−9.8T ↓ v =9.8T
A B
For same speed 29.4−9.8T =9.8T
so T = 1.5
and V = 14.7
H =29.4×1.5−0.5×9.8×1.52
+0. ×9.8×1.52
= 44.1
Method 2
V2 =29.42−2×9.8×x=2×9.8×(H−x)
29.42 =19.6H so H = 44.1
Relative velocity is 29.4 so
44.1
T =
29.4
Using v = u + at
V =0+9.8×1.5 = 14.7 | M1
A1
M1
E1
F1
M1
A1
M1
B1
A1
M1
E1
M1
F1 | Either attempted. Allow sign errors and g = 9.81
etc
Both correct
Attempt to equate. Accept sign errors and T = 1.5
substituted in both.
If 2 subs there must be a statement about equality
FT T or V, whichever is found second
Sum of the distance travelled by each attempted
cao
Attempts at V2 for each particle equated. Allow
sign errors, 9.81 etc
Allow h , h without h =H−h
1 2 1 2
Both correct. Require h =H−h but not an
1 2
equation.
cao
Any method that leads to T or V
Any method leading to the other variable
Other approaches possible. If ‘clever’ ways seen,
reward according to weighting above. | 7
7
Small stones A and B are initially in the positions shown in Fig. 6 with B a height $H$ m directly above A.
\includegraphics{figure_5}
At the instant when B is released from rest, A is projected vertically upwards with a speed of $29.4\text{ms}^{-1}$. Air resistance may be neglected.
The stones collide $T$ seconds after they begin to move. At this instant they have the same speed, $V\text{ms}^{-1}$, and A is still rising.
By considering when the speed of A upwards is the same as the speed of B downwards, or otherwise, show that $T = 1.5$ and find the values of $V$ and $H$. [7]
\hfill \mbox{\textit{OCR MEI M1 Q5 [7]}}