CBSE Class 11 Physics Exemplar NCERT Solutions: Chapter 4 - Laws of Motion

Solution: Uniform translatory motion implies that every single point on the object is moving with the exact same velocity. This means the velocity has a constant magnitude and a constant direction. Therefore, all parts of the ball must have the same velocity, and this velocity must be constant over time. Option (a) is incorrect as motion is involved. Option (b) is insufficient as it doesn't specify that *all parts* have the same velocity. Option (d) is incorrect because translatory motion does not necessarily involve spinning.

Answer: (c)

Solution: According to Newton's second law of motion, the net force acting on an object is equal to the rate of change of its momentum ($F\; =\; \backslash frac\{dp\}\{dt\}$). If the metre scale is moving with uniform velocity, its momentum ($p\; =\; mv$) is constant because its velocity ($v$) is constant. Therefore, the change in momentum ($\backslash Delta\; p$) is zero, which means the net external force ($F$) acting on the scale must be zero.

Furthermore, if the entire scale is moving with uniform velocity, it means that each part of the scale is also moving with the same uniform velocity. For a net force of zero to result in uniform velocity (and not rotation), the net torque about any point, including the center of mass, must also be zero. If there were a net torque, the scale would start rotating or change its rotational state.

Answer: (b)

Solution: The mass of the cricket ball is given as $m\; =\; 150\; \backslash text\{\; g\}$, which needs to be converted to kilograms: $m\; =\; 0.15\; \backslash text\{\; kg\}$.

The initial velocity is $\backslash mathbf\{u\}\; =\; (3\backslash hat\{i\}\; +\; 4\backslash hat\{j\})\; \backslash ,\backslash text\{m/s\}$ and the final velocity is $\backslash mathbf\{v\}\; =\; -(3\backslash hat\{i\}\; +\; 4\backslash hat\{j\})\backslash text\{\; m/s\}$.

The change in momentum ($\backslash Delta\; \backslash mathbf\{p\}$) is defined as the final momentum minus the initial momentum:

$\backslash Delta\; \backslash mathbf\{p\}\; =\; m\backslash mathbf\{v\}\; -\; m\backslash mathbf\{u\}\; =\; m(\backslash mathbf\{v\}\; -\; \backslash mathbf\{u\})$

Substitute the given values:

$\backslash Delta\; \backslash mathbf\{p\}\; =\; (0.15\; \backslash text\{\; kg\})\; [-(3\backslash hat\{i\}\; +\; 4\backslash hat\{j\})\; \backslash text\{\; m/s\}\; -\; (3\backslash hat\{i\}\; +\; 4\backslash hat\{j\})\; \backslash text\{\; m/s\}]$

Combine the terms inside the bracket:

$\backslash Delta\; \backslash mathbf\{p\}\; =\; (0.15)\; [(-3\; -\; 3)\backslash hat\{i\}\; +\; (-4\; -\; 4)\backslash hat\{j\}]\; \backslash text\{\; kg\; m/s\}$

$\backslash Delta\; \backslash mathbf\{p\}\; =\; (0.15)\; [-6\backslash hat\{i\}\; -\; 8\backslash hat\{j\}]\; \backslash text\{\; kg\; m/s\}$

Now, multiply by the mass:

$\backslash Delta\; \backslash mathbf\{p\}\; =\; -(0.15\; \backslash times\; 6)\backslash hat\{i\}\; -\; (0.15\; \backslash times\; 8)\backslash hat\{j\}\; \backslash text\{\; kg\; m/s\}$

$\backslash Delta\; \backslash mathbf\{p\}\; =\; -(0.9\backslash hat\{i\}\; +\; 1.2\backslash hat\{j\})\; \backslash text\{\; kg\; m/s\}$

Thus, the change in momentum is $-(0.9\backslash hat\{i\}\; +\; 1.2\backslash hat\{j\})\; \backslash text\{\; kg\; m/s\}$.

Answer: (c)

Solution: From the previous problem, the change in momentum ($\backslash Delta\; \backslash mathbf\{p\}$) was found to be $\backslash Delta\; \backslash mathbf\{p\}\; =\; -(0.9\backslash hat\{i\}\; +\; 1.2\backslash hat\{j\})\; \backslash text\{\; kg\; m/s\}$.

The magnitude of the momentum transferred is the magnitude of this change in momentum vector. We calculate the magnitude using the Pythagorean theorem for vectors:

$|\backslash Delta\; \backslash mathbf\{p\}|\; =\; \backslash sqrt\{(\backslash Delta\; p\_x)^2\; +\; (\backslash Delta\; p\_y)^2\}$

Here, $\backslash Delta\; p\_x\; =\; -0.9$ and $\backslash Delta\; p\_y\; =\; -1.2$.

$|\backslash Delta\; \backslash mathbf\{p\}|\; =\; \backslash sqrt\{(-0.9)^2\; +\; (-1.2)^2\}\; \backslash text\{\; kg\; m/s\}$

$|\backslash Delta\; \backslash mathbf\{p\}|\; =\; \backslash sqrt\{0.81\; +\; 1.44\}\; \backslash text\{\; kg\; m/s\}$

$|\backslash Delta\; \backslash mathbf\{p\}|\; =\; \backslash sqrt\{2.25\}\; \backslash text\{\; kg\; m/s\}$

$|\backslash Delta\; \backslash mathbf\{p\}|\; =\; 1.5\; \backslash text\{\; kg\; m/s\}$

Therefore, the magnitude of the momentum transferred is $1.5\; \backslash text\{\; kg\; m/s\}$.

Answer: (c)

Solution: The principle of conservation of momentum states that the total momentum of an isolated system remains constant. This principle can be derived from Newton's laws of motion.

According to Newton's second law, the net external force acting on a system is equal to the rate of change of its total momentum ($F\_\{\backslash text\{ext\}\}\; =\; \backslash frac\{dp\}\{dt\}$). If the net external force is zero ($F\_\{\backslash text\{ext\}\}\; =\; 0$), then the rate of change of momentum is zero ($\backslash frac\{dp\}\{dt\}\; =\; 0$), which implies that the momentum ($p$) is constant ($p\; =\; \backslash text\{constant\}$).

In a collision between particles within a system, the forces exchanged between the particles are internal forces. According to Newton's third law, these forces occur in equal and opposite pairs. Therefore, the vector sum of these internal forces acting on the system is always zero. If there are no external forces acting on the system, then the total momentum of the system is conserved.

Thus, conservation of momentum is understood from both Newton's second law (which relates force to momentum change) and Newton's third law (which ensures internal forces cancel out, leading to zero net external force if no external forces are present).

Answer: (d)