A body of mass 10 kg explodes into three masses in the ratio 2 : 3 : 5. Two light fragment fly off at right angle with velocity 12m/ s and 8m/s. Find the velocity of third fragment.

After explosion,

Let 2 kg mass after explosion fly off along X-axis and 3 kg mass fly off along Y-axis perpendicular to direction of motion of 2 kg mass.

Suppose the velocity of third fragment after explosion is <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre>

∴ Momentum of 2 kg mass is, 

             stack straight P subscript 1 with rightwards arrow on top space equals space straight m subscript 1 stack straight v subscript 1 with rightwards arrow on top space equals space 2 cross times 12 straight i with hat on top space equals space 24 straight i with hat on top space space Ns 

Momentum of 3 kg mass is, 

            stack straight P subscript 2 with rightwards arrow on top space equals space straight m subscript 2 stack straight v subscript 2 with rightwards arrow on top space equals space 3 cross times 8 straight j with hat on top space equals space 24 space straight j with hat on top space space Ns 

Momentum of 5 kg mass is, 

          stack straight P subscript 3 with rightwards arrow on top space equals space straight m subscript 3 space stack straight v subscript 3 with rightwards arrow on top space equals space 3 cross times straight v with rightwards arrow on top space equals space 3 straight v with rightwards arrow on top space space Ns 

According to the law of conservation of linear momentum, 

              stack straight P subscript 1 with rightwards arrow on top plus stack straight P subscript 2 with rightwards arrow on top plus stack straight P subscript 3 with rightwards arrow on top space equals space 0 

rightwards double arrow         24 straight i with hat on top space plus space 24 straight j with hat on top space plus space 5 straight v with rightwards arrow on top space equals space 0 

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#6 {main}</pre> 

∴          
space space space open vertical bar straight v with rightwards arrow on top close vertical bar space equals space square root of left parenthesis 4.8 right parenthesis squared plus left parenthesis 4.8 right parenthesis squared end root                       

                  equals space 4.8 square root of 2 space straight m divided by straight s   

This is the required velocity of the third fragment. 

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State and prove law of conservation of momentum.

Law of conservation of momentum:
It states that total momentum of system remains conserved in the absence of external force.

Law of conservation of momentum:It states that total momentum of syst


Law of conservation of momentum:It states that total momentum of syst

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Force-time graph for a body is shown in figure. Locate the regions of time if the body has uniform motion, uniformly accelerated motion, non-uniformly accelerated motion, uniformly retarded motion, non-uniformly retarded motion.


Ans.

(i) For <pre>uncaught exception: <b>mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied</b><br /><br />in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre>,

The force acting on the body is constant. Thus,  the body has uniform accelerated motion.

(ii) Forspace space space 5 less than straight t less or equal than 8,

Force on the body is zero, therefore body has uniform motion. 

(iii) For 8 less than straight t less or equal than 13

The force is negative and increases linearly with time in magnitude, thus the motion is non-uniformly retarded motion.

(iv) For 13 less than straight t less or equal than 17,

The force is positive and increases linearly with time, thus the motion is non-uniformly accelerated motion.

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#6 {main}</pre>,

The force acting on the body is constant, thus the body has uniform accelerated motion.

(vi) For 20 less than straight t less or equal than 24,

The force acting on the body is constant and negative, thus the body has uniformly retarded motion.

566 Views

Show that Newton’s second law is the real law of motion.

Second law is the real law of motion. This can be proved by showing that first law and third law are contained in second law of motion.
(i) First law is contained in second law. According to the second law of motion,

                                     straight F with rightwards arrow on top space equals space straight m straight a with rightwards arrow on top
space space therefore                               straight F with rightwards arrow on top space equals space 0 space space space space space space rightwards double arrow space space space space straight a with rightwards arrow on top space equals space 0
That is if no force is acting on the body then its acceleration is zero, means if a body is at rest, it remains at rest and if the body is moving in straight line with constant velocity, it continues to do so. This is what the first law states. Hence first law is contained in the second law.
(ii) Third law is contained in second. According to the second law of motion,

Second law is the real law of motion. This can be proved by showing t
        

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A man weighing M kg, stands on a weighing machine inside a lift. What will be the reading of the machine if it:
(a) descends with acceleration a,
(b) ascends with an acceleration a and
(c) moves with constant velocity in upward direction?


(a)  

The true weight of body = Mg

Reaction of floor = R

Net force acting on the body is,

                     F = Mg  - R

If a is the acceleration with which lift moves downward, then

                     F = M a

∴                Ma = Mg - R

i.e.,               R = M(g - a)

Thus the apparent weight of the body is,

                 straight W subscript straight a space equals space straight R space equals space straight M left parenthesis straight g minus straight a right parenthesis

(b)

If lift is ascending in upward direction with acceleration a then

                 Ma = R - Mg

or               straight R space equals space straight W subscript straight a space equals space straight M left parenthesis straight g plus straight a right parenthesis 

(c)  

If lift moves with constant velocity then acceleration, a = 0. 

∴                straight W subscript straight a space equals space Mg

So, reading on the machine = Mg

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