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Nuetrik [128]
4 years ago
13

define momentum of a body.state law of co?servation of momentum.prove with the help of third law of motion that total momentum o

f two bodies is conserved during collision provided no external force acts.
Physics
1 answer:
xxTIMURxx [149]4 years ago
4 0
The momentum of a body is defined as the product of its mass and velocity....

law of momentum of conservation depends on collision,
for a collision occurring between object 1 and object 2 in an isolated system,the total momentum of the two objects before the Collision is equal to the total momentum of two objects after the Collision.....

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The law of conservation of momentum states that the total momentum of interacting objects does not _______. This means the total
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6 0
1 year ago
The uniform rods AB and BC weigh 24 ky and kg, respectively,and the small wheel at C is of negligible weight. If the wheel ismov
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The velocity of pin B after rod AB has rotated through 90* is vb = 3.2549 m/s.

<h3>What is Potential and Kinetic energy?</h3>

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mass of rod, mab = 2.4kg

mass of rod, mbc = 4kg

conservation of energy

T_{1}  + V_{1} = T_{2}  + V_{2}

h_{ab}  = h_{bc}  = 0.18m

potential energy at position 1,

V1 = m_{ab} gh_{ab}  + m_{bc} gh_{bc}

V1 = 2.5 * 9.81 * 0.18 + 4 * 9.81 * 0.18

V1 = 11.30112

kinetic energy T1 at position 1 is zero

potential energy at position 2 is zero

K.E at position 2,

T_{2} = \frac{1}{2} l_{ab} w^{2}_{ab} +  \frac{1}{2} m_{bc} v^{2}_{G} +  \frac{1}{2} lw^{2}_{bc}

l_{ab} =\frac{m_{ab} l^{2}_{ab}  }{3}

= 1/3 *4 * (0.36)²

=0.10368kg m²

l =\frac{m_{bc} l^{2}_{bc}  }{12}

= 1/12 *4 * (0.6)²

=0.12kg m²

on putting the values in above equation we get,

T₂ = 1.0667vb²

0 + 11.30112 = 1.0667vb² + 0

vb = 3.2549 m/s

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Some rewards are 2.33 miles in a hour so you have to move in 700 degrees to get the system moving faster soo 700+ 2.33 divide by 3
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