The bombardment!

We know that the Momentum formula is $p = p'$ Because they are stuck together at the first, so we can write $p = (m_{1} + m_{2}) v$ . After they breaks into two the momentum become $p' = m_{1}v_{1}' + m_{2}v_{2}'$ . So, the formula become $(m_{1} + m_{2}) v = m_{1}v_{1}' + m_{2}v_{2}'$ And we can fill with the following information from the question. Hence, it will be $28 \times 13.5 = 21 \times 9 + 7 \times v_{2}'$ It is positive because it doesn't say whether it is bounce back or not. So we can easily solve the problem $378 = 189 + 7v_{2}'$ $378 - 189 = 7v_{2}'$ $189= 7v_{2}'$ $v_{2}' = \frac{189}{7}$ $v_{2}' = 27$ So, the velocity of the other piece will be $27 \frac{m}{s}$ .

mass of whole body at initial=28kg velocity " " " " " " " " =13.5m/s

mass of larger body after breaks up=21 velocity of larger body =9m/s

mass of smaller body after breaks up=28-21=7kg velocity of smaller body " " " "' " " = ?

By the law of conservation of linear momentum =>M V=m1 v1+m2*v2 =>28 x 13.5=21 x 7+7 x v2 =>7xv2=378-189 =>v2= 189/7 v2=27.Ans

Using law of conservation of momentum, Initial momentum =final momentum or, 28X13.5= 21X9+(28-21)Xv' ( because total weight is 28kg and weight of 1 part is 21kg, let velocity of 2nd part be v') or, 378= 189+7v' or, 7v'= 189 Therefore, v'= 27 Ans.

Thanks for the explanation @ @Jansen Wu

arriving the same formula and answer :) @Jansen Wu Good job man :)