Q: A bullet of mass 10 g moving horizontal with a velocity of 400 m/s strikes a wood block of mass 2 kg which is suspended by light inextensible string of length 5 m. As a result, the centre of gravity of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges horizontally from the block will be

120 m/s. By conservation of momentum p_i=p_f: (0.01)\times400+0=2v+(0.01)v' ...(i). The block's speed just after collision is v=\sqrt{2gh}=\sqrt{2\times10\times0.1}=\sqrt2 m/s ...(ii). Solving (i) and (ii) gives v'\approx120 m/s.

Q: A balloon with mass m is descending down with an acceleration a (where, a<g). How much mass should be removed from it so that it starts moving up with an acceleration a?

\dfrac{2ma}{g+a}. Descending: mg-B=ma ...(i), where B is the (unchanged) buoyant force. After removing mass, the new mass m' moves up: B-m'g=m'a ...(ii). Adding, mg-m'g=ma+m'a\Rightarrow m(g-a)=m'(g+a)\Rightarrow m'=\dfrac{m(g-a)}{g+a}. Mass removed =m-m'=m\left[1-\dfrac{g-a}{g+a}\right]=\dfrac{2ma}{g+a}.

Q: An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass 1 kg moves with a speed of 12 ms^{-1} and the second part of mass 2 kg moves with 8 ms^{-1} speed. If the third part flies off with 4 ms^{-1} speed, then its mass is

5 kg. Momentum is conserved: \mathbf p_1+\mathbf p_2+\mathbf p_3=0. Taking the two perpendicular parts along \hat i,\hat j: 1\times12\hat i+2\times8\hat j+\mathbf p_3=0\Rightarrow\mathbf p_3=-(12\hat i+16\hat j), so |\mathbf p_3|=\sqrt{12^2+16^2}=\sqrt{144+256}=20 kg-m/s. Then m_3=\dfrac{p_3}{v_3}=\dfrac{20}{4}=5 kg.

Q: Two spheres A and B of masses m_1 and m_2 respectively collide. A is at rest initially and B is moving with velocity v along x-axis. After collision, B has a velocity \dfrac{v}{2} in a direction perpendicular to the original direction. The mass A moves after collision in the direction

\theta=\tan^{-1}\left(\dfrac{1}{2}\right) to the x-axis. By conservation of momentum, \mathbf p_i=m_2 v\hat i and \mathbf p_f=m_2\dfrac{v}{2}\hat j+m_1\mathbf v_1. Equating gives \mathbf v_1=\dfrac{m_2}{m_1}v\hat i+\dfrac{m_2}{m_1}\dfrac{v}{2}\hat j, so \tan\theta=\dfrac{y}{x}=\dfrac12\Rightarrow\theta=\tan^{-1}\left(\dfrac12\right) to the x-axis.

Q: A man of 50 kg mass is standing in a gravity free space at a height of 10 m above the floor. He throws a stone of 0.5 kg mass downwards with a speed 2 ms^{-1}. When the stone reaches the floor, the distance of the man above the floor will be

10.1 m. The centre of mass stays fixed, so m_1r_1=m_2r_2\Rightarrow r_2=\dfrac{m_1r_1}{m_2}=\dfrac{0.5\times10}{50}=0.1 m (the man rises 0.1 m while the stone falls 10 m). Hence the man's distance above the floor =10+0.1=10.1 m.

Q: A body, under the action of a force \mathbf F=6\hat i-8\hat j+10\hat k, acquires an acceleration of 1 ms^{-2}. The mass of this body must be

10\sqrt2 kg. |\mathbf F|=\sqrt{6^2+8^2+10^2}=\sqrt{36+64+100}=10\sqrt2 N. Since F=ma with a=1 ms^{-2}, m=\dfrac{10\sqrt2}{1}=10\sqrt2 kg.

Question 1

A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is nearly (g=10 m/s^2)

Accepted Answer

4.2 kg-m/s. Impulse =m|\Delta\bar v| where \Delta\bar v=v_2-v_1. Velocity just before striking v_1=-\sqrt{2gh}=-\sqrt{2	imes10	imes10}=-10\sqrt2 m/s and after rebound v_2=\sqrt{2gh}=10\sqrt2 m/s. So |Impulse|=0.15\,|10\sqrt2-(-10\sqrt2)|=0.15	imes2	imes10\sqrt2\approx4.2 kg-m/s.

Question 2

A truck is stationary and has a bob suspended by a light string, in a frame attached to the truck. The truck, suddenly moves to the right with an acceleration of a. The pendulum will tilt

Accepted Answer

to the left and angle of inclination of the pendulum with the vertical is 	an^{-1}\left(\dfrac{a}{g}ight). As the truck moves to the right, the bob swings to the left due to inertia, experiencing a pseudo acceleration a. In the truck's frame the string makes an angle 	heta with the vertical where 	an	heta=\dfrac{ma}{mg}\Rightarrow	heta=	an^{-1}\left(\dfrac{a}{g}ight).

Question 3

A particle moving with velocity \mathbf v is acted by three forces shown by the vector triangle PQR. The velocity of the particle will

Accepted Answer

remain constant. Three forces represented by the three sides of a triangle taken in order are in equilibrium, so \mathbf F_{net}=\mathbf F_{PQ}+\mathbf F_{QR}+\mathbf F_{RP}=0. Then \mathbf F_{net}=m\mathbf a=m\dfrac{d\mathbf v}{dt}=0\Rightarrow\dfrac{d\mathbf v}{dt}=0, i.e. \mathbf v= constant. So the velocity of the particle remains constant.

Question 4

A rigid ball of mass m strikes a rigid wall at 60^\circ and gets reflected without loss of speed as shown in the figure. The value of impulse imparted by the wall on the ball will be

Accepted Answer

mv. Impulse equals the change in the perpendicular component of momentum. J=\Delta p=mv_f-mv_i=mv\cos60^\circ-(-mv\cos60^\circ)=2mv\cos60^\circ=2mv	imes\dfrac12=mv.

Question 5

A bullet of mass 10 g moving horizontal with a velocity of 400 m/s strikes a wood block of mass 2 kg which is suspended by light inextensible string of length 5 m. As a result, the centre of gravity of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges horizontally from the block will be

Accepted Answer

120 m/s. By conservation of momentum p_i=p_f: (0.01)	imes400+0=2v+(0.01)v' ...(i). The block's speed just after collision is v=\sqrt{2gh}=\sqrt{2	imes10	imes0.1}=\sqrt2 m/s ...(ii). Solving (i) and (ii) gives v'\approx120 m/s.

Question 6

A balloon with mass m is descending down with an acceleration a (where, a<g). How much mass should be removed from it so that it starts moving up with an acceleration a?

Accepted Answer

\dfrac{2ma}{g+a}. Descending: mg-B=ma ...(i), where B is the (unchanged) buoyant force. After removing mass, the new mass m' moves up: B-m'g=m'a ...(ii). Adding, mg-m'g=ma+m'a\Rightarrow m(g-a)=m'(g+a)\Rightarrow m'=\dfrac{m(g-a)}{g+a}. Mass removed =m-m'=m\left[1-\dfrac{g-a}{g+a}\right]=\dfrac{2ma}{g+a}.

Question 7

An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass 1 kg moves with a speed of 12 ms^{-1} and the second part of mass 2 kg moves with 8 ms^{-1} speed. If the third part flies off with 4 ms^{-1} speed, then its mass is

Accepted Answer

5 kg. Momentum is conserved: \mathbf p_1+\mathbf p_2+\mathbf p_3=0. Taking the two perpendicular parts along \hat i,\hat j: 1	imes12\hat i+2	imes8\hat j+\mathbf p_3=0\Rightarrow\mathbf p_3=-(12\hat i+16\hat j), so |\mathbf p_3|=\sqrt{12^2+16^2}=\sqrt{144+256}=20 kg-m/s. Then m_3=\dfrac{p_3}{v_3}=\dfrac{20}{4}=5 kg.

Question 8

Two spheres A and B of masses m_1 and m_2 respectively collide. A is at rest initially and B is moving with velocity v along x-axis. After collision, B has a velocity \dfrac{v}{2} in a direction perpendicular to the original direction. The mass A moves after collision in the direction

Accepted Answer

heta=	an^{-1}\left(\dfrac{1}{2}ight) to the x-axis. By conservation of momentum, \mathbf p_i=m_2 v\hat i and \mathbf p_f=m_2\dfrac{v}{2}\hat j+m_1\mathbf v_1. Equating gives \mathbf v_1=\dfrac{m_2}{m_1}v\hat i+\dfrac{m_2}{m_1}\dfrac{v}{2}\hat j, so 	an	heta=\dfrac{y}{x}=\dfrac12\Rightarrow	heta=	an^{-1}\left(\dfrac12ight) to the x-axis.

Question 9

A man of 50 kg mass is standing in a gravity free space at a height of 10 m above the floor. He throws a stone of 0.5 kg mass downwards with a speed 2 ms^{-1}. When the stone reaches the floor, the distance of the man above the floor will be

Accepted Answer

10.1 m. The centre of mass stays fixed, so m_1r_1=m_2r_2\Rightarrow r_2=\dfrac{m_1r_1}{m_2}=\dfrac{0.5	imes10}{50}=0.1 m (the man rises 0.1 m while the stone falls 10 m). Hence the man's distance above the floor =10+0.1=10.1 m.

Question 10

A body, under the action of a force \mathbf F=6\hat i-8\hat j+10\hat k, acquires an acceleration of 1 ms^{-2}. The mass of this body must be

Accepted Answer

10\sqrt2 kg. |\mathbf F|=\sqrt{6^2+8^2+10^2}=\sqrt{36+64+100}=10\sqrt2 N. Since F=ma with a=1 ms^{-2}, m=\dfrac{10\sqrt2}{1}=10\sqrt2 kg.

Question 11

A 0.5 kg ball moving with a speed of 12 m/s strikes a hard wall at an angle of 30^\circ with the wall. It is reflected with the same speed and at the same angle. If the ball is in contact with the wall for 0.25 s, the average force acting on the wall is

Accepted Answer

24 N. Only the component of momentum perpendicular to the wall reverses, so the change in momentum is 2mv\sin30^\circ. Given m=0.5 kg, v=12 m/s, t=0.25 s, the average force F=\dfrac{2mv\sin30^\circ}{t}=\dfrac{2	imes0.5	imes12	imes\sin30^\circ}{0.25}=24 N.

Question 12

A block of mass m is placed on a smooth wedge of inclination 	heta. The whole system is accelerated horizontally, so that the block does not slip on the wedge. The force exerted by the wedge on the block (g is acceleration due to gravity) will be

Accepted Answer

\dfrac{mg\sin	heta}{\cos	heta}. Let the wedge be accelerated towards the left. In the wedge's non-inertial frame the block experiences a pseudo force towards the right. For the block not to slip on the smooth incline, mg\sin	heta=ma_{pseudo}\cos	heta\Rightarrow a_{pseudo}=\dfrac{g\sin	heta}{\cos	heta}, giving the required force as \dfrac{mg\sin	heta}{\cos	heta}.

Question 13

An object of mass 3 kg is at rest. If a force \mathbf{F}=(6t^2\hat{i}+4t\hat{j}) N is applied on the object, then the velocity of the object at t=3 s is

Accepted Answer

18\hat{i}+6\hat{j}. By Newton's second law \mathbf{F}=m\dfrac{d\mathbf{v}}{dt}, so d\mathbf{v}=\dfrac{1}{3}(6t^2\hat{i}+4t\hat{j})\,dt. Integrating from 0 to 3 s, \mathbf{v}=\dfrac{1}{3}\left[2t^3\hat{i}+2t^2\hat{j}\right]_0^3=\dfrac{1}{3}(54\hat{i}+18\hat{j})=18\hat{i}+6\hat{j}.

Question 14

A player takes 0.1 s in catching a ball of mass 150 g moving with velocity of 20 m/s. The force imparted by the ball on the hands of the player is

Accepted Answer

30 N. Force imparted equals the rate of change of momentum, F=\dfrac{m(v_1-v_2)}{\Delta t}. With m=0.150 kg, v_1=20 m/s, v_2=0 and \Delta t=0.1 s, F=\dfrac{0.150	imes(20-0)}{0.1}=30 N.

Question 15

1 kg body explodes into three fragments. The ratio of their masses is 1:1:3. The fragments of same mass move perpendicular to each other with speeds 30 m/s, while the heavier part remains in the initial direction. The speed of heavier part is

Accepted Answer

10\sqrt{2} m/s. The masses are 0.2, 0.2 and 0.6 kg. The two equal fragments move perpendicular to each other with 30 m/s, so their combined momentum is \sqrt{(0.2	imes30)^2+(0.2	imes30)^2}=6\sqrt{2} kg-m/s. The heavier part must carry equal and opposite momentum, so v=\dfrac{6\sqrt{2}}{0.6}=10\sqrt{2} m/s.

Question 16

A particle of mass 1 kg is thrown vertically upwards with speed 100 m/s. After 5 s, it explodes into two parts. One part of mass 400 g comes back with speed 25 m/s, what is the speed of other part just after explosion?

Accepted Answer

100 m/s upwards. After 5 s the velocity is v=u-gt=100-10	imes5=50 m/s (upwards). By conservation of momentum Mv=m_1v_1+m_2v_2 with M=1 kg, m_1=0.4 kg, v_1=-25 m/s, m_2=0.6 kg: 1	imes50=0.4	imes(-25)+0.6\,v_2\Rightarrow 50=-10+0.6\,v_2\Rightarrow v_2=100 m/s upwards.

Question 17

A ball of mass 3 kg moving with a speed of 100 m/s, strikes a wall at an angle 60^\circ (as shown in figure). The ball rebounds at the same speed and remains in contact with the wall for 0.2 s, the force exerted by the ball on the wall is

Accepted Answer

1500\sqrt{3} N. Only the component perpendicular to the wall reverses, so \Delta P=2mv\sin60^\circ=2	imes3	imes100	imes\dfrac{\sqrt{3}}{2}=300\sqrt{3} kg-m/s. The force F=\dfrac{\Delta P}{t}=\dfrac{300\sqrt{3}}{0.2}=1500\sqrt{3} N.

Question 18

The force on a rocket moving with a velocity 300 m/s is 210 N. The rate of consumption of fuel of rocket is

Accepted Answer

0.7 kg/s. Thrust force F_t=v_r\left(-\dfrac{dm}{dt}\right), so the rate of consumption of fuel -\dfrac{dm}{dt}=\dfrac{F_t}{v_r}=\dfrac{210}{300}=0.7 kg/s.

Question 19

A 5000 kg rocket is set for vertical firing. The exhaust speed is 800 ms^{-1}. To give an initial upward acceleration of 20 m/s^2, the amount of gas ejected per second to supply the needed thrust will be (g=10 ms^{-2})

Accepted Answer

187.5 kg s^{-1}. The thrust must supply the weight plus the force for the acceleration, so \left(-\dfrac{dm}{dt}ight)=\dfrac{m(g+a)}{v_r}=\dfrac{5000(10+20)}{800}=\dfrac{5000	imes30}{800}=187.5 kg s^{-1}.

Question 20

A bullet is fired from a gun. The force on the bullet is given by F=600-2	imes10^5t where F is in newton and t in second. The force on the bullet becomes zero as soon as it leaves the barrel. What is the average impulse imparted to the bullet?

Accepted Answer

0.9 N-s. The force is zero when 600-2	imes10^5t=0\Rightarrow t=3	imes10^{-3} s. Impulse =\int_0^t F\,dt=\left[600t-\dfrac{2	imes10^5t^2}{2}ight]_0^{3	imes10^{-3}}=1.8-0.9=0.9 N-s.

Laws of Motion — NEET Physics PYQ MCQs with Solutions

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