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Question: A body of mass 2 kg moving with a velocity \(\hat i + 2\hat j - 3\hat km/s\) collides with another b...

A body of mass 2 kg moving with a velocity i^+2j^−3k^m/s\hat i + 2\hat j - 3\hat km/s collides with another body of mass 3 kg moving with a velocity 2i^+j^+k^m/s2\hat i + \hat j + \hat km/s. If they stick together the velocity in m/sm/s of the composite body is
A. 15(8i^+7j^−3k^)m/s\dfrac{1}{5}(8\hat i + 7\hat j - 3\hat k)m/s
B. 15(−4i^+1j^−3k^)m/s\dfrac{1}{5}( - 4\hat i + 1\hat j - 3\hat k)m/s
C. 15(5i^+1j^−1k^)m/s\dfrac{1}{5}(5\hat i + 1\hat j - 1\hat k)m/s
D. 15(−4i^+8j^−3k^)m/s\dfrac{1}{5}( - 4\hat i + 8\hat j - 3\hat k)m/s

Explanation

Solution

Hint: In extreme cases of inelastic collision, the masses stick together. In an inelastic collision the kinetic energy is not conserved, only the total energy is conserved. Almost all macroscopic collisions are inelastic collisions. Remember that after sticking together the composite body move with one velocity vf{v_f} and the mass will be the sum of both i.e. m1+m2{m_1} + {m_2}.

Formula used:
P=m×vP = m \times v, where PP denotes the momentum of the body, mm denotes the mass and vv denotes its velocity.
M×vf=m1×v1+m2×v2M \times {v_f} = {m_1} \times {v_1} + {m_2} \times {v_2}, where MM denotes the total or combined mass, vf{v_f} denotes the final velocity of the composite body, m1{m_1} and m2{m_2} denote the masses of the first and second body and v1{v_1} and v2{v_2} shows their velocities.

Complete step by step answer:
This is an inelastic collision; therefore there is no conservation of kinetic energy. But the total energy as well as the momentum of the system is conserved. We can use the conservation of momentum to solve this question i.e. the sum of momentum of the two bodies which collided will be equal to the momentum of the composite body. This is represented by the equation M×vf=m1×v1+m2×v2M \times {v_f} = {m_1} \times {v_1} + {m_2} \times {v_2}, where MM denotes the total or combined mass, vf{v_f} denotes the final velocity of the composite body, m1{m_1} and m2{m_2} denote the masses of the first and second body and v1{v_1} and v2{v_2} shows their velocities.
It is given that m1=2Kg{m_1} = 2Kg , v1=i^+2j^−3k^m/s{v_1} = \hat i + 2\hat j - 3\hat km/s, m2=3Kg{m_2} = 3Kg, v2=2i^+j^+k^m/s{v_2} = 2\hat i + \hat j + \hat km/s
MMwill the total mass i.e. the sum of two masses M=5KgM = 5Kg
Substituting these values in the equation M×vf=m1×v1+m2×v2M \times {v_f} = {m_1} \times {v_1} + {m_2} \times {v_2} we get,
5×vf=2×(i^+2j^−3k^)+3×(2i^+j^+k^)5 \times {v_f} = 2 \times (\hat i + 2\hat j - 3\hat k) + 3 \times (2\hat i + \hat j + \hat k)
5×vf=(2i^+4j^−6k^)+(6i^+3j^+3k^)5 \times {v_f} = (2\hat i + 4\hat j - 6\hat k) + (6\hat i + 3\hat j + 3\hat k)
5×vf=(2+6)i^+(4+3)j^+(−6+3)k^5 \times {v_f} = (2 + 6)\hat i + (4 + 3)\hat j + ( - 6 + 3)\hat k
vf=15((8)i^+(7)j^+(−3)k^){v_f} = \dfrac{1}{5}((8)\hat i + (7)\hat j + ( - 3)\hat k)
Thus we can say that the correct option is A.

Note: Normally all ordinary collisions are considered to be inelastic. This is because for a collision to be considered as elastic its kinetic energy as well as its momentum/energy should be conserved. The kinetic energy is not conserved in ordinary collisions due to the action of internal friction. In most natural cases some of the kinetic energy is converted to other forms. For example in some cases, the kinetic energy is converted into vibrational energy of the atom, in some others as sound energy etc.