physics ncert solutions class 11th

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation – particle is projected from the point O.

Let time taken in reaching from point O to point P is T.

for journey O to P

y=0,u_y= V_osin $\beta$ ,a_y= -gcos $\alpha ,t=T$

y=u_yt + $\frac{1}{2}{a}_y{t}^2$

0= V_osin $\beta T+\frac{1}{2}\left(-gcos\alpha \right)T^2$

T[V_osin $\beta -\frac{gcos\alpha }{2}$ T]=0

T = time of flight = $\frac{2V_{o}sin\beta }{gcos\alpha }$

 

Motion along OX

x= L ,u_x= V_ocos $\beta$ , a_x= -gsin $\alpha$

t =T = $\frac{2V_{o}sin\beta }{gcos\alpha }$

x= u_xt+ $\frac{1}{2}{a}_x{t}^2$

L= V₀cos $\beta T$ + $\frac{1}{2}(-gsin\alpha )T^2$

L= T[V₀cos $\beta -\frac{1}{2}gsin\alpha T$ ]

L=  $\frac{2V_{o}sin\beta }{gcos\alpha }$ [V_ocos $\beta -\frac{V_{o}sin\alpha sin\beta }{cos\alpha }$ ]

L= $\frac{2v_o^{2}sin\beta }{g{cos}^2\alpha }\mathrm{c}\mathrm{o}\mathrm{s}?\left(\alpha +\beta \right)$

Z= sin $\beta \mathrm{c}\mathrm{o}\mathrm{s}?\left(\alpha +\beta \right)$

 = sin $\beta [cos\alpha cos\beta -sin\alpha sin\beta ]$

= $\frac{1}{2}(cos\alpha +sin2\beta -2sin\alpha .{sin}^2\beta )$

= ½ [sin2] $\beta cos\alpha -sin\alpha (1-cos2\beta )$

= $\frac{1}{2}[sin2\beta cos\alpha +cos2\beta .sin\alpha -sin\alpha ]$

= $\frac{1}{2}$  [sin(2 $\beta +\alpha$ )-sin $\alpha$ ]

For z maximum

2 $\beta +\alpha =\frac{\pi }{2}$  , $\beta =\frac{\pi }{4}-\frac{\alpha }{2}$

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation – target T is at horizontal distance x= R+ $?x$ and between point of projection y= -h

Maximum horizontal range R= $\frac{v_o^2}{g}\theta =45$ …………1

Horizontal component of initial velocity = V_ocos $\theta$

Vertical component of initial velocity = -V_osin $\theta$

So h = (-V_osin $\theta$ )t + $\frac{1}{2}gt$ ²………….2

R+ $?x$  = V_ocos $\theta *t$

So t= $\frac{R+?x}{v_{o}cos\theta }$

Substituting value of t in 2 we get

So h = (-V₀sin $\theta$ ) $\left(\mathrm{}\frac{R+?x}{v_{o}cos\theta }\right)+\frac{1}{2}g({\mathrm{}\frac{R+?x}{v_{o}cos\theta })}^2$

H = -(R+ $?x$ )tan $\theta$ + $\frac{1}{2}g\left(\frac{({R+?x)}^2}{v_o^2{cos}^2\theta }\right)$

$\theta =45$ ,  h = -(R+ $?x$ )tan $45$ + $\frac{1}{2}g\left(\frac{({R+?x)}^2}{v_o^2{cos}^{2}45}\right)$

So h = -(R+ $?x$ ) $1$ + $\frac{1}{2}g\left(\frac{({R+?x)}^2}{v_o^2\left(\frac{1}{2}\right)}\right)$

So h = -(R+ $?x$ )+ $\frac{{(R+?x)}^2}{R}$

So h = -R- $?x$ +(R+ $\frac{{?x}^2}{R}+2?x$ )

 h= $?x+\frac{{?x}^2}{R}$

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation- speed of jackets = 125m/s

Height of hill = 500m

To cross the hill vertical component of velocity should be grater than this value u_y= $\sqrt[]{2gh}$

$=\sqrt[]{2*10*500}=100m/s$

So u²= u_x²+u_y²

Horizontal component of initial velocity u_x = $\sqrt[]{u^2-u_y^2}=\sqrt[]{{125}^2-{100}^2}=75m/s$

Time taken to reach the top of hill t= $\sqrt[]{\frac{2h}{g}}=\sqrt[]{\frac{2*500}{10}}=10s$

Time taken to reach the ground in 10 sec = 75 (10)= 750m

Distance through which the canon has to be moved =800-750=50m

Speed with which canon can move = 2m/s

Time taken canon = 50/2= 25s

Total time t= 25+10+10= 45s

This is a multiple choice answer as classified in NCERT Exemplar

This is a multiple choice answer as classified in NCERT Exemplar

(b, d) When a man of mass m climbs up the staircases of height L, work done by the gravitational force on the man is mgl work done by internal muscular forces will be mgL as the change in kinetic is almost zero.

Hence total work done =-mgL + mgL=0

As the point of application of the contact forces does not move hence work done by reaction forces will be zero.

This is a multiple choice answer as classified in NCERT Exemplar

(c) m =150g =3/20kg

Time of contact =0.001s

U=126km/h= $\frac{126*1000}{60*60}=\frac{35m}{s}$

V= -35m/s

Change in momentum of the ball = m (v-u)= $\frac{3}{20}\left(-35-35\right)kgm/s$

=21/2

F= dp/dt=- $\frac{21/2}{0.001}N$ = - 1.05 $*{10}^{4}N$

Here – negative sign indicates that force will be opposite to the direction of movement of the ball before hitting.

This is a multiple choice answer as classified in NCERT Exemplar

(b) First velocity of the iron sphere increases and after sometimes becomes constant called terminal velocity. Hence according first KE increases and then becomes constant which is best represented by b.

This is a multiple choice answer as classified in NCERT Exemplar

(d) h =15m, v= 1m/s, m= 10kg, g= 10m/s²

From conservation of mechanical energy

(PE+KE)_initial= (PE+KE)_final

Mgh + $\frac{1}{2}m{v}^2$ =0+KE

KE= mgh + $\frac{1}{2}m{v}^2$

KE= 10 $*10*1.5+\frac{1}{2}*10*1$

KE= 150+5=155J