Moderate -0.3 This is a straightforward SUVAT question requiring standard application of kinematic equations for vertical motion. Part (i) is routine calculation, part (ii) involves simple rearrangement, and part (iii) requires setting up equations for equal heights and comparing velocities—all standard M1 techniques with no novel insight needed. Slightly easier than average due to the guided structure and familiar context.
5 A particle \(P\) is projected vertically upwards, from horizontal ground, with speed \(8.4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
Show that the greatest height above the ground reached by P is 3.6 m .
A particle Q is projected vertically upwards, from a point 2 m above the ground, with speed \(\mathrm { um } \mathrm { s } ^ { - 1 }\). The greatest height abovetheground reached by Q is also 3.6 m .
Find the value of \(u\).
It is given that P and Q are projected simultaneously.
Show that, at the instant when P and Q are at the same height, the particles have the same speed and are moving in opposite directions.
\(v_P^2 = 8.4^2 - 2g(s+/2)\) and \(v_Q^2 = v_Q^2 = \frac{v_Q^2 = 5.6^2 - 2g\lbrack(s+/2)\rbrack}{\text{the speeds are always the same at the same heights.}}\)
A1
5(iii) cont
\(0 = 8.4 - gt\) and \(0 = 5.6 - gt\)
A1
5(iii) cont
\(t_P = 6/7\) and \(t_Q = 4/7\) means there is a time interval when Q has started to descend but P is still rising, and there will be a position where they have the same height but are moving in opposite directions.
A1
5(i) | $8.4^2 − 2gs_{max} = 0$ | M1 | Using $v^2 = u^2 +/- 2gs$ with $v = 0$ or $u = 0$
5(i) | Height is 3.6m | A1 |
5(i) | | (AG) | A1
5(i) | | [3] |
5(ii) | | M1 | Using $u^2 = +/- 2g(\text{ans}(i) - 2)$
5(ii) | $u = 5.6$ | A1 |
5(ii) | | [2] |
5(iii) | EITHER (time when at same height) | M1 | Using $s = ut + \frac{1}{2}at$ for P and for Q, $a = -/+g$, expressions for s terms must differ
5(iii) | $s+/2 = 8.4t − \frac{1}{2}gt^2$ and $(s+/2) = 5.6t − \frac{1}{2}gt^2$ | A1 | Or 8.4 $(-\frac{1}{2}gt^2)$+$/2$ Correct sign for g, cv(5,6), +/-2 in only one equation cao
5(iii) | $t = 5/7$ (0.714) | A1 |
5(iii) | $v_P = 8.4 - 0.714g$ and $v_Q = 5.6 - 0.714g$ | M1 | Using $v = u+at$ for P and for Q, $a = +/-g$, cv(t) Correct sign for g, cv(5,6), candidates answer for t (including sign) cao
5(iii) | $v_P = 1.4$ and $v_Q = -1.4$ | A1 |
5(iii) | | [6] |
5(iii) | OR (time when at same speed in opposite directions) | M1 | Using $v = u+at$ for P and for Q, $a = +/-g$
5(iii) | $v = 8.4 - gt$ and $-v = 5.6 - gt$ | A1 | Correct sign for g, cv(5,6)
5(iii) | $v = 1.4$ [or $t = 5/7$ (0.714)] | A1 | Only one correct answer is needed
5(iii) | (with $v = 1.4$) $1.4^2 = 8.4^2 - 2gs_p$ and $(-1.4)^2 = 5.6^2 - 2gs_Q$ | M1 | Using $v^2 = u^2 + 2as$ for P and for Q, $a = +/-g$, cv(5,6)
5(iii) | $s_P = 3.5$ and $s_Q = 1.5$ | A1 | Correct sign for g, cv(5,6), candidate's answer for v (including - for Q) cao
5(iii) | $s = 8.4x0.714 − \frac{1}{2}gx0.714^2$ and $s = 5.6x0.714 − \frac{1}{2}gx0.714^2$ | M1 | Using $s = ut + \frac{1}{2}at^2$ for P and for Q, $a = +/-g$, cv(t)
5(iii) | $s_P = 3.5$ and $s_Q = 1.5$ | A1 | Correct sign for g, cv(5,6), candidate's answer for t (including sign of t if negative) cao]
5(iii) | OR (motion related to greatest height and verification) | M1 | Using $v = u+at$ for P and for Q, $a = +/-g$
5(iii) | $0 = 8.4 - gt$ and $0 = 5.6 - gt$ | A1 | Both values correct mid-interval $t = (6/7+4/7)/2 = 0.714$ {Or semi-interval = $6/7-4/7)/2=1/7$}
5(iii) | $t_P = 6/7$ and $t_Q = 4/7$ | A1 | cao
5(iii) | $v_P = 8.4 - 0.714g$ and $v_Q = 5.6 - 0.714g$ | M1 | Using $v^2 = u^2 + 2as$ for P and for Q, $a = +/-g$, cv(5,6) and cv(t)
5(iii) | $s_P = 8.4x0.714 − \frac{1}{2}gx0.714^2$ and $s_Q = 5.6x0.714 − \frac{1}{2}gx0.714^2$ | A1 | Correct sign for g, cv(5,6), candidate's answer for t (including sign of t if negative) cao]
5(iii) | $s_P = 3.5$ $s_Q = 1.5$ $\{s_P = 0/7 − \frac{1}{2}g(1/7)^2$ and $s_Q = 0/7 + \frac{1}{2}g(1/7)^2\}$ | A1 | cao
5(iii) | | [6] |
5(iii) | OR (without finding exactly where or when) | M1 | Using $v^2 = u^2 + 2as$ for P and for Q, $a = +/-g$, cv(5,6), different expressions for s. Correct sign for g, cv(5,6), (s+/2) used only once cao. Verbal explanation essential
5(iii) cont | $v_P^2 = 8.4^2 - 2g(s+/2)$ and $v_Q^2 = v_Q^2 = \frac{v_Q^2 = 5.6^2 - 2g\lbrack(s+/2)\rbrack}{\text{the speeds are always the same at the same heights.}}$ | A1 | Using $v = u+at$ for P and for Q, $a = +/-g$ Correct sign for g, correct choice for velocity zero, cv(5,6)
5(iii) cont | $0 = 8.4 - gt$ and $0 = 5.6 - gt$ | A1 |
5(iii) cont | $t_P = 6/7$ and $t_Q = 4/7$ means there is a time interval when Q has started to descend but P is still rising, and there will be a position where they have the same height but are moving in opposite directions. | A1 | cao. Verbal explanation essential
5 A particle $P$ is projected vertically upwards, from horizontal ground, with speed $8.4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Show that the greatest height above the ground reached by P is 3.6 m .
A particle Q is projected vertically upwards, from a point 2 m above the ground, with speed $\mathrm { um } \mathrm { s } ^ { - 1 }$. The greatest height abovetheground reached by Q is also 3.6 m .\\
(ii) Find the value of $u$.
It is given that P and Q are projected simultaneously.\\
(iii) Show that, at the instant when P and Q are at the same height, the particles have the same speed and are moving in opposite directions.
\hfill \mbox{\textit{OCR M1 2007 Q5 [11]}}