Two objects of masses $200g$ and $500g$ possess velocities $10\hat im/s$ and $3\hat i + 5\hat j$ m/s respectively. The velocity of their centre of mass in m/s is
a. $5\hat i - 25\hat j$
b. $\dfrac{5}{7}\hat i - 25\hat j$
c. $5\hat i - \dfrac{{25}}{7}\hat j$
d. $25\hat i - \dfrac{5}{7}\hat j$
Answer
Verified
448.5k+ views
Hint: To describe its motion, we consider a point in the body where the entire mass of the body is supposed to concentrate to describe its motion is called center of mass. Motion of the body is represented by the path of the particle at the center of mass point. Using the above data apply the velocity of mass formula.
Complete step by step answer:
The total momentum of the body is conserved when the initial momentum is equal to the final momentum of a system.
To describe its motion, we consider a point in the body where the entire mass of the body is supposed to concentrate to describe its motion is called center of mass.
Motion of the body is represented by the path of the particle at the center of mass point.
The center of mass is located at the centroid when the rigid body is with uniform density. The center of mass for a disc which is uniform would be at a center.
In some cases, the center of mass may not fall on the object. For a ring the center of mass is located at its center.
Let us consider two blocks A and B
${M_A} = 200g = 0.2kg$ and ${M_B} = 500g = 0.5kg$
Initial momentum is given by
$ \Rightarrow {M_A}{V_A} + {M_B}{V_B} = 0.2 \times 10\hat i + 0.5 \times \left( {3\hat i + 5\hat j} \right)$
$ \Rightarrow {M_i} = 3.5\hat i + 2.5\hat j$
Then the final momentum
$ \Rightarrow {M_{total}}{V_{cm}} = 0.7{V_{cm}}$
Then from conservation of momentum,
$ \Rightarrow {M_i} = 3.5i + 2.5j = 0.7{V_{cm}}$
Hence velocity of center of mass,
\[ \Rightarrow {V_{cm}} = 5\hat i + \dfrac{{25}}{7}\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{j} \]
Hence, the correct answer is option (C).
Note: By vector addition, we can determine the center of mass of an object. If the particle moves in uniform velocity then the magnitude of the center of mass is obtained by parallelogram law of vectors. The center of mass is located at the centroid when the rigid body is with uniform density. The center of mass for a disc which is uniform would be at a center.
Complete step by step answer:
The total momentum of the body is conserved when the initial momentum is equal to the final momentum of a system.
To describe its motion, we consider a point in the body where the entire mass of the body is supposed to concentrate to describe its motion is called center of mass.
Motion of the body is represented by the path of the particle at the center of mass point.
The center of mass is located at the centroid when the rigid body is with uniform density. The center of mass for a disc which is uniform would be at a center.
In some cases, the center of mass may not fall on the object. For a ring the center of mass is located at its center.
Let us consider two blocks A and B
${M_A} = 200g = 0.2kg$ and ${M_B} = 500g = 0.5kg$
Initial momentum is given by
$ \Rightarrow {M_A}{V_A} + {M_B}{V_B} = 0.2 \times 10\hat i + 0.5 \times \left( {3\hat i + 5\hat j} \right)$
$ \Rightarrow {M_i} = 3.5\hat i + 2.5\hat j$
Then the final momentum
$ \Rightarrow {M_{total}}{V_{cm}} = 0.7{V_{cm}}$
Then from conservation of momentum,
$ \Rightarrow {M_i} = 3.5i + 2.5j = 0.7{V_{cm}}$
Hence velocity of center of mass,
\[ \Rightarrow {V_{cm}} = 5\hat i + \dfrac{{25}}{7}\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{j} \]
Hence, the correct answer is option (C).
Note: By vector addition, we can determine the center of mass of an object. If the particle moves in uniform velocity then the magnitude of the center of mass is obtained by parallelogram law of vectors. The center of mass is located at the centroid when the rigid body is with uniform density. The center of mass for a disc which is uniform would be at a center.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success
Master Class 12 English: Engaging Questions & Answers for Success
Master Class 12 Economics: Engaging Questions & Answers for Success
Master Class 12 Chemistry: Engaging Questions & Answers for Success
Master Class 12 Social Science: Engaging Questions & Answers for Success
Class 12 Question and Answer - Your Ultimate Solutions Guide
Trending doubts
10 examples of friction in our daily life
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
State and prove Bernoullis theorem class 11 physics CBSE
Pigmented layer in the eye is called as a Cornea b class 11 biology CBSE
State the laws of reflection of light