 # [Physics Momentum Question] Hii, Could anyone help to solve this question?

### Physics Question

4.Tiga blok dengan jisim 10 \mathrm{~kg}, 5 \mathrm{~kg} dan 1 \mathrm{~kg}
masing-masing diletakkan di atas meja yang licin, seperti dalam rajah di bawah. Pada mulanya, blok berjisim 10 \mathrm{~kg} bergerak dengan halaju 10 \mathrm{~m} \mathrm{~s}^{-1} \mathrm{ke} arah kanan. Dengan menganggap semua perlanggaran adalah kenyal, apakah halaju akhir jisim I \mathrm{~kg} selepas perlanggaran?

Three blocks with masses 10 \mathrm{~kg} 5 \mathrm{~kg} and 1 \mathrm{~kg} respectively are placed on a frictionless tabletop as shown in the diagram below. Initially, the 10 \mathrm{~kg} block is moving with 10 \mathrm{~m} \mathrm{~s}^{-1} to the right. Assume all the collisions are elastic, what is the final velocity of the 1 \mathrm{~kg} mass after collision?

A. 11.1 m s^{-1}
B. 22.2 ms^{-1}
C. 33.3 m s^{-1}
D. 44.4 ms^{-1}

### Physics Solution

Notice that the question mentions “assume all collisions are elastic”, what do we know about an elastic collision?

1. total momentum before and after collision are conserved
2. total kinetic energy before and after collision are conserved. Using the formula
m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2
we get
10(10) + 5(0) = 10v_1 + 5v_2
100 = 10v_1 + 5v_2
20 = 2v_1 + v_2
v_2=20-2v_1 ------ (1)
Since we cannot use one equation to solve for 2 unknowns, we have to use the energy conservation formula.
KE before collision = KE after collision
0.5(10)(10)^2 = 0.5(10)v_1^2 + 0.5(5)v_2^2
500 = 5v_1^2 + 2.5v_2^2
200=2v_1^2 + v_2^2 ------(2)

Sub (1) into (2)
200=2v_1^2 + (20-2v_1)^2
6v_1^2 -80v_1 + 200 = 0
3v_1^2 -40v_1 + 100 = 0
(3v_1 -10) ( v_1-10) = 0
We reject v_1=10 because this would mean that the 5kg block would remain stationary and no KE is transferred to the 5kg block. It doesn’t make sense, right?

v_1= \frac{10}{3}, v_2=\frac{40}{3}

We don’t look at v_1 anymore because there would be no further collision between the 10kg and 5kg block as v_1<v_2
So the second collision would be between the 5kg block and the 1kg block. We apply the same steps we used previously to solve:

5( \frac{40}{3}) + 1(0) =5v_3 + 1v_4
200=15v_3 + 3v_4
v_3= \frac{200-3v_4}{15} ------ (3)

0.5(5)( \frac{40}{3})^2= 0.5(5)v_3^2 + 0.5(1)v_4^2
\frac{4000}{9}=\frac{5}{2}v_3^2+\frac{1}{2}v_4^2
8000=45v_3^2 + 9v_4^2 ------(4)

Substitute (3) into (4) ,
8000=45 (\frac{200-3v_4}{15})^2 + 9v_4^2
8000 = \frac{45}{225}(40000 -1200v_4 +9v_4^2) +9v_4^2
54v_4^2-1200v_4=0
v_4(54v_4-1200)=0
We reject v_4=0 because some kinetic energy must be transferred to the 1kg block.

v_4=\frac{1200}{54}=22.2ms^{-1}
This is quite a long winded question so I hope you understand the working, if you have any doubts do feel free to clarify! Thank you so much  