Standard +0.3 This is a straightforward momentum and energy problem requiring conservation of momentum to find m, then calculating KE loss for two cases (same/opposite directions after collision). The algebra is simple and the method is standard for M1, making it slightly easier than average but not trivial due to the two-case consideration.
4 Two small smooth spheres \(A\) and \(B\), of equal radii and of masses 4 kg and \(m \mathrm {~kg}\) respectively, lie on a smooth horizontal plane. Initially, sphere \(B\) is at rest and \(A\) is moving towards \(B\) with speed \(6 \mathrm {~ms} ^ { - 1 }\). After the collision \(A\) moves with speed \(1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and \(B\) moves with speed \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
Find the two possible values of the loss of kinetic energy due to the collision.
\(KE = \frac{1}{2} \times m \times v^2\); FT \(4.5m\) for \(\text{KE}_B\)
or \(\text{KE}_A\) after \(= \frac{1}{2} \times 4 \times 1.5^2\) \((4.5\text{ J})\)
or \(\text{KE}_B\) after \(= \frac{1}{2} \times 6 \times 3^2\) \((27\text{ J})\)
or \(\text{KE}_B\) after \(= \frac{1}{2} \times 10 \times 3^2\) \((45\text{ J})\)
KE loss \(= [\frac{1}{2}\times4\times6^2 - \frac{1}{2}\times4\times1.5^2 - \frac{1}{2}\times6\times3^2]\) or \([\frac{1}{2}\times4\times6^2 - \frac{1}{2}\times4\times1.5^2 - \frac{1}{2}\times10\times3^2]\)
M1
Uses KE loss = KE before \(-\) KE after
Loss of KE \(= 40.5\text{ J}\) or \(22.5\text{ J}\)
A1
## Question 4:
| Answer | Marks | Guidance |
|--------|-------|----------|
| For using conservation of momentum (either case) | M1 | |
| $6 \times 4 = 3m + 4 \times 1.5$ or $6 \times 4 = 3m - 4 \times 1.5$ | A1 | |
| $m = 6$ and $m = 10$ | A1 | |
| $\text{KE}_A$ initial $= \frac{1}{2} \times 4 \times 6^2$ $(72\text{ J})$ | B1 FT | $KE = \frac{1}{2} \times m \times v^2$; FT $4.5m$ for $\text{KE}_B$ |
| or $\text{KE}_A$ after $= \frac{1}{2} \times 4 \times 1.5^2$ $(4.5\text{ J})$ | | |
| or $\text{KE}_B$ after $= \frac{1}{2} \times 6 \times 3^2$ $(27\text{ J})$ | | |
| or $\text{KE}_B$ after $= \frac{1}{2} \times 10 \times 3^2$ $(45\text{ J})$ | | |
| KE loss $= [\frac{1}{2}\times4\times6^2 - \frac{1}{2}\times4\times1.5^2 - \frac{1}{2}\times6\times3^2]$ or $[\frac{1}{2}\times4\times6^2 - \frac{1}{2}\times4\times1.5^2 - \frac{1}{2}\times10\times3^2]$ | M1 | Uses KE loss = KE before $-$ KE after |
| Loss of KE $= 40.5\text{ J}$ or $22.5\text{ J}$ | A1 | |
4 Two small smooth spheres $A$ and $B$, of equal radii and of masses 4 kg and $m \mathrm {~kg}$ respectively, lie on a smooth horizontal plane. Initially, sphere $B$ is at rest and $A$ is moving towards $B$ with speed $6 \mathrm {~ms} ^ { - 1 }$. After the collision $A$ moves with speed $1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $B$ moves with speed $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
Find the two possible values of the loss of kinetic energy due to the collision.\\
\hfill \mbox{\textit{CAIE M1 2020 Q4 [6]}}