363549
Statement A : The moment after a stone is released out of an accelerated train, there is no horizontal force or acceleration on the stone. Statement B : Force on a body at a given time is determined by the situation at the location of the body at that time.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
Both statements are are correct. So option (3) is correct.
PHXI05:LAWS OF MOTION
363550
An object of mass \(10\,kg\) moves at a constant speed of \(10\,m/s.\) A constant force that acts for \(4 s\) on the object, gives it a speed \(2\,m/s\) in opposite direction. The force acting on the object is
1 \(-3 N\)
2 \(-30 N\)
3 \(3 N\)
4 \(30 N\)
Explanation:
From the first equation of motion, \(v = u + at\) \( \Rightarrow v - u = at\) \( \Rightarrow a = \frac{{v - u}}{t}\) Given, \(u = 10\;m/s,\quad v = - 2\;m/s\) and \(t=4 s\). \(a = \frac{{ - 2 - 10}}{4} = - 3\;m/{s^2}\) So, force acting on the object \(F = ma = 10 \times ( - 3) = - 30\;N\)
PHXI05:LAWS OF MOTION
363551
Figure shows (\(x\), \(t\)), (\(y\), \(t\)) diagram of a particle moving in 2-dimensions. If the particle has a mass of 500 \(g\), the force acting on the particle is
1 0.5 \(N\) along \(x\) - axis
2 0.5 \(N\) along \(y\) - axis
3 0.1 \(N\) along \(y\) - axis
4 0.1 \(N\) along \(x\) - axis
Explanation:
Since the graphs between \(x\) and \(t\) is a straight line and passing through the origin. At \(x = 2,\;t = 2\) \(\therefore \,\,\,x = t\) At \(t = 2,\;y = 4\) Since the graph between \(y\) and \(t\) is a parabola. \(\therefore \,\,\,\,y = {t^2}\) \(\therefore \,\,\,\,{v_x} = \frac{{dx}}{{dt}} = 1\) and \({a_x} = \frac{{d{v_x}}}{{dt}} = 0\) and \({v_y} = \frac{{dy}}{{dt}} = 2t\) and \({a_y} = 2\,m{s^{ - 2}}\) The force acting on the particle is \(F = m{a_y} = (0.5\,kg)\,(2\,m{s^{ - 2}})\) \( = 1N\) along \(y\) -axis
PHXI05:LAWS OF MOTION
363552
A body of mass 5 \(kg\) starts from the origin with an initial velocity \(\vec u = 30\,\hat i + 40\,\hat j\,m{s^{ - 1}}.\) If a constant force \(\vec F = - (\hat i + 5\,\hat j)N\) acts on the body, the time in which the \(y\)-component of the velocity becomes zero is
363549
Statement A : The moment after a stone is released out of an accelerated train, there is no horizontal force or acceleration on the stone. Statement B : Force on a body at a given time is determined by the situation at the location of the body at that time.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
Both statements are are correct. So option (3) is correct.
PHXI05:LAWS OF MOTION
363550
An object of mass \(10\,kg\) moves at a constant speed of \(10\,m/s.\) A constant force that acts for \(4 s\) on the object, gives it a speed \(2\,m/s\) in opposite direction. The force acting on the object is
1 \(-3 N\)
2 \(-30 N\)
3 \(3 N\)
4 \(30 N\)
Explanation:
From the first equation of motion, \(v = u + at\) \( \Rightarrow v - u = at\) \( \Rightarrow a = \frac{{v - u}}{t}\) Given, \(u = 10\;m/s,\quad v = - 2\;m/s\) and \(t=4 s\). \(a = \frac{{ - 2 - 10}}{4} = - 3\;m/{s^2}\) So, force acting on the object \(F = ma = 10 \times ( - 3) = - 30\;N\)
PHXI05:LAWS OF MOTION
363551
Figure shows (\(x\), \(t\)), (\(y\), \(t\)) diagram of a particle moving in 2-dimensions. If the particle has a mass of 500 \(g\), the force acting on the particle is
1 0.5 \(N\) along \(x\) - axis
2 0.5 \(N\) along \(y\) - axis
3 0.1 \(N\) along \(y\) - axis
4 0.1 \(N\) along \(x\) - axis
Explanation:
Since the graphs between \(x\) and \(t\) is a straight line and passing through the origin. At \(x = 2,\;t = 2\) \(\therefore \,\,\,x = t\) At \(t = 2,\;y = 4\) Since the graph between \(y\) and \(t\) is a parabola. \(\therefore \,\,\,\,y = {t^2}\) \(\therefore \,\,\,\,{v_x} = \frac{{dx}}{{dt}} = 1\) and \({a_x} = \frac{{d{v_x}}}{{dt}} = 0\) and \({v_y} = \frac{{dy}}{{dt}} = 2t\) and \({a_y} = 2\,m{s^{ - 2}}\) The force acting on the particle is \(F = m{a_y} = (0.5\,kg)\,(2\,m{s^{ - 2}})\) \( = 1N\) along \(y\) -axis
PHXI05:LAWS OF MOTION
363552
A body of mass 5 \(kg\) starts from the origin with an initial velocity \(\vec u = 30\,\hat i + 40\,\hat j\,m{s^{ - 1}}.\) If a constant force \(\vec F = - (\hat i + 5\,\hat j)N\) acts on the body, the time in which the \(y\)-component of the velocity becomes zero is
363549
Statement A : The moment after a stone is released out of an accelerated train, there is no horizontal force or acceleration on the stone. Statement B : Force on a body at a given time is determined by the situation at the location of the body at that time.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
Both statements are are correct. So option (3) is correct.
PHXI05:LAWS OF MOTION
363550
An object of mass \(10\,kg\) moves at a constant speed of \(10\,m/s.\) A constant force that acts for \(4 s\) on the object, gives it a speed \(2\,m/s\) in opposite direction. The force acting on the object is
1 \(-3 N\)
2 \(-30 N\)
3 \(3 N\)
4 \(30 N\)
Explanation:
From the first equation of motion, \(v = u + at\) \( \Rightarrow v - u = at\) \( \Rightarrow a = \frac{{v - u}}{t}\) Given, \(u = 10\;m/s,\quad v = - 2\;m/s\) and \(t=4 s\). \(a = \frac{{ - 2 - 10}}{4} = - 3\;m/{s^2}\) So, force acting on the object \(F = ma = 10 \times ( - 3) = - 30\;N\)
PHXI05:LAWS OF MOTION
363551
Figure shows (\(x\), \(t\)), (\(y\), \(t\)) diagram of a particle moving in 2-dimensions. If the particle has a mass of 500 \(g\), the force acting on the particle is
1 0.5 \(N\) along \(x\) - axis
2 0.5 \(N\) along \(y\) - axis
3 0.1 \(N\) along \(y\) - axis
4 0.1 \(N\) along \(x\) - axis
Explanation:
Since the graphs between \(x\) and \(t\) is a straight line and passing through the origin. At \(x = 2,\;t = 2\) \(\therefore \,\,\,x = t\) At \(t = 2,\;y = 4\) Since the graph between \(y\) and \(t\) is a parabola. \(\therefore \,\,\,\,y = {t^2}\) \(\therefore \,\,\,\,{v_x} = \frac{{dx}}{{dt}} = 1\) and \({a_x} = \frac{{d{v_x}}}{{dt}} = 0\) and \({v_y} = \frac{{dy}}{{dt}} = 2t\) and \({a_y} = 2\,m{s^{ - 2}}\) The force acting on the particle is \(F = m{a_y} = (0.5\,kg)\,(2\,m{s^{ - 2}})\) \( = 1N\) along \(y\) -axis
PHXI05:LAWS OF MOTION
363552
A body of mass 5 \(kg\) starts from the origin with an initial velocity \(\vec u = 30\,\hat i + 40\,\hat j\,m{s^{ - 1}}.\) If a constant force \(\vec F = - (\hat i + 5\,\hat j)N\) acts on the body, the time in which the \(y\)-component of the velocity becomes zero is
363549
Statement A : The moment after a stone is released out of an accelerated train, there is no horizontal force or acceleration on the stone. Statement B : Force on a body at a given time is determined by the situation at the location of the body at that time.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both Statements are correct.
4 Both Statements are incorrect.
Explanation:
Both statements are are correct. So option (3) is correct.
PHXI05:LAWS OF MOTION
363550
An object of mass \(10\,kg\) moves at a constant speed of \(10\,m/s.\) A constant force that acts for \(4 s\) on the object, gives it a speed \(2\,m/s\) in opposite direction. The force acting on the object is
1 \(-3 N\)
2 \(-30 N\)
3 \(3 N\)
4 \(30 N\)
Explanation:
From the first equation of motion, \(v = u + at\) \( \Rightarrow v - u = at\) \( \Rightarrow a = \frac{{v - u}}{t}\) Given, \(u = 10\;m/s,\quad v = - 2\;m/s\) and \(t=4 s\). \(a = \frac{{ - 2 - 10}}{4} = - 3\;m/{s^2}\) So, force acting on the object \(F = ma = 10 \times ( - 3) = - 30\;N\)
PHXI05:LAWS OF MOTION
363551
Figure shows (\(x\), \(t\)), (\(y\), \(t\)) diagram of a particle moving in 2-dimensions. If the particle has a mass of 500 \(g\), the force acting on the particle is
1 0.5 \(N\) along \(x\) - axis
2 0.5 \(N\) along \(y\) - axis
3 0.1 \(N\) along \(y\) - axis
4 0.1 \(N\) along \(x\) - axis
Explanation:
Since the graphs between \(x\) and \(t\) is a straight line and passing through the origin. At \(x = 2,\;t = 2\) \(\therefore \,\,\,x = t\) At \(t = 2,\;y = 4\) Since the graph between \(y\) and \(t\) is a parabola. \(\therefore \,\,\,\,y = {t^2}\) \(\therefore \,\,\,\,{v_x} = \frac{{dx}}{{dt}} = 1\) and \({a_x} = \frac{{d{v_x}}}{{dt}} = 0\) and \({v_y} = \frac{{dy}}{{dt}} = 2t\) and \({a_y} = 2\,m{s^{ - 2}}\) The force acting on the particle is \(F = m{a_y} = (0.5\,kg)\,(2\,m{s^{ - 2}})\) \( = 1N\) along \(y\) -axis
PHXI05:LAWS OF MOTION
363552
A body of mass 5 \(kg\) starts from the origin with an initial velocity \(\vec u = 30\,\hat i + 40\,\hat j\,m{s^{ - 1}}.\) If a constant force \(\vec F = - (\hat i + 5\,\hat j)N\) acts on the body, the time in which the \(y\)-component of the velocity becomes zero is
