Laws of Motion Test - 13

If all the particles of the body move with the same velocity in same straight line, then the motion is called uniform motion or uniform translatory motion.

As the metre scale is moving with uniform velocity.

\(\therefore\) No change in its velocity i.e., acceleration of it is zero by Newton’s second law

\(\vec F_{net}\) = m x 0 = 0

\(\therefore\) Hence net or resultant force must acting on body is zero

\(\therefore\) \(\vec \tau = \vec r \times \vec F\)

So torque must be zero. Hence, for uniform motion force and torque, both must be zero.So, It verifies the option (b).

m = 150 g = 0.15 kg

\(\vec u =(3\hat i +4\hat j)ms^{-1}\) and \(\vec v= -(3\hat i+4\hat j)ms^{-1}\)

Change in momentum \(\Delta \vec p\) Final momentum - Initial momentum = \(m\vec v-m\vec u\)

= \(m[\vec v-\vec u]=0.15[-(3\hat i+4\hat j)-(3\hat i+4\hat j)]\)

= \(0.15[-3\hat i-4\hat j-3\hat i-4\hat j]\)

= \(0.15[-6\hat i-8\hat j]\)

\(\Delta \vec p=-0.9\hat i-1.2\hat j=-[0.9\hat i+1.2\hat j]\)

Hence, this verifies the option(c).

m = 150 g = 0.15 kg

\(\vec u =(3\hat i +4\hat j)ms^{-1}\) and \(\vec v= -(3\hat i+4\hat j)ms^{-1}\)

Change in momentum \(\Delta \vec p\) Final momentum - Initial momentum = \(m\vec v-m\vec u\)

= \(m[\vec v-\vec u]=0.15[-(3\hat i+4\hat j)-(3\hat i+4\hat j)]\)

= \(0.15[-3\hat i-4\hat j-3\hat i-4\hat j]\)

= \(0.15[-6\hat i-8\hat j]\)

\(\Delta \vec p=-0.9\hat i-1.2\hat j=-[0.9\hat i+1.2\hat j]\)

Magnitude = \(|\Delta \vec p|\) = \(\sqrt{(0.9)^2+(1.2)^2}=\sqrt{0.81+1.44}=\sqrt{2.25}\)

= \(1.5\,kg\,ms^{-1}\)

So, this verifies the option (c).

(i) By Newton’s second law \({d\vec p\over dt}=\vec F_{ext}\)

As \(\vec F_{ext}\) in law of conservation of momentum is zero.

i.e., \(\vec F_{ext}\) = 0

\({d\vec p\over dt}\) = 0

⇒ \(\vec p\) is constant.

(ii) By Newton’s third law action force is equal to reaction force in magnitude but in opposite direction.

\(\therefore\) \(\vec F_{12}=-\vec F_{21}\,\,\,\,(\vec F_{ext}=0)\)

\({d\vec p_{12}\over dt}={-d\vec p_{21}\over dt}\,or\, d\vec p_{12}=-d\vec p_{21}\)

\(d\vec p_{12}+d\vec p_{21}=0\)

So proves the law of conservation of momentum and verifies the option (d).

\(\vec F=m\vec a = m.{d^2x\over dt^2}\)

\(\because\) x(t) = pt + \(qt^2\) + \(rt^3\), \(p=3ms^{-1}\), \(q=4ms^{-2}\), \(r=5ms^{-3}\)

x(t) = 3t + \(4t^2+5t^3\)

\({dx(t)\over dt}=3+8t+15t^2\)

\({d^2x(t)\over dt^2}=0+8+30t\)

\(\Big[{d^2x(t)\over dt^2}\Big]_{t=2}=8+30\times 2=68ms^{-2}\)

\(\therefore\,\,\vec F\) = 2 x 68 = 136 N.

Verifies the option (a).

\(\vec u=(6\hat i-12\hat j)ms^{-1}\)

\(F=(-3\hat i+4\hat j)N\)

m = 5 kg    \(\vec a={\vec F\over m}=\Big({-3\over5}\hat i+{4\over5}\hat j\Big)ms^{-2}\)

As the final velocity has only Y component and X component is zero

\(v_x=u_x+a_xt\)

\(0=6+{-3\over 5}t \implies {3\over 5}t=6\)

t = 10sec. Verifies the option (b).

u = 0, \(\vec v=vi\), t = 2, m = m

v = u + at

\(\vec v\hat i=0+\vec a \times 2\)

\(\vec a={\vec v\hat i\over 2}\)

\(\vec F=m\vec a={m\vec v\over2}\hat i\)

Force by the engine is an internal force.

Hence, the force \({m\vec v\over2}\hat i\) acting on a car is due to force of friction, which moves the car in the eastward direction.

\(At\,\, t={1\over8}s\,\,a(t)=-16\pi^2Asin 4\pi {1\over8}\)

a(t) = \(-16\pi ^2 Asin{\pi\over 2}\)

a(t) = -16\(\pi^2\)A

\(\therefore\) Force at t = \(1\over 8\) sec,

F = ma = m(-16\(\pi^2\)A)

F = -16\(\pi^2\)Am N

verifies the option (a).

(b) Impulse = Change in momentum between t = 0 sec and t = \(1\over4\)sec

From (i) F(t) varies from zero at t = 0 sec to the maximum value at

i.e., F(t) = -16\(\pi^2\)mA at t = \(1\over 8\)sec,

Hence by \(\vec I=\vec Ft\,\,\,\,at\,\,\,\,t={1\over4}s\)

we get \(\vec I=-16\pi^2mA\times {1\over4}=-4\pi^2mA\)

So, It verified the option (b).

(d) Mass = m

x(t) = 0 for t < 0

x(t) = A sin \(4\pi t\)    for 0 < t < \(({1\over4})s\)

x(t) = 0       for t > \(({1\over4})s\)

for 0 < t < \(({1\over4})s\)   x(t) = A sin \(4\pi t\)

\(v={dx(t)\over dt}=4A\pi cos4\pi t\)

\(a= {dv\over dt}=-16\pi^2mAsin4\pi t\)

As the force is a function of time as in above equation, so force acting on the particle is not constant verifies option (d).

Mass of each ball m = 0.005 kg

Speed of each ball v = 5 m/s

\(\therefore\) Initial momentum of each ball \(\vec p_i\) = \(m\vec v\)

\(\vec p_i\) = (0.005)(5) = 0.25 kg \(ms^{-1}\)

= 0.25 N-s  (1)

As after the collision, the direction of the velocity of each ball is reversed on rebounding

\(\therefore\) Final momentum of each ball \(\vec p\) = \(m(-\vec v)\)

\(\vec p_f\) = 0.005 x (-5) = -0.25 kg \(ms^{-1}\)

= -0.25 N-s

Eqns (1) and (2) verifies option (d).

\(\therefore\) Impulse imparted to each ball = Change in momentum of each ball

= \(p_f-p_i\)

= -0.25 - (0.25) = -0.50 kg \(ms^{-1}\)

= -0.50 N-s

Equation (3) verifies option (c).

i.e., the magnitude of impulse imparted by one ball due to collision with the other ball = 0.50 kg \(ms^{-1}\) . These two impulses are opposite to each other.