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BlackZzzverrR [31]
3 years ago
7

Assume the ground is uniformly level. If the horizontal component a projectile's velocity is doubled, but the vertical component

is unchanged, what is the effect on the time of flight
Physics
1 answer:
Katarina [22]3 years ago
3 0

Answer:

Explanation:

Time of flight = 2 x u sinα / g where u sinα is vertical component of projectile's velocity u .

So Time of flight = 2 x vertical component / g

vertical component = constant

g is also constant so

Time of flight will also be constant .

It will remain unchanged .

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a 3.0 kg mass moving to the right at 1.4 m/s collides in a perfectly inelastic collision with a 2.0 kg mass initially at rest. w
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The velocity of the combined mass after the collision is 0.84 ms-1.

<u>Explanation:</u>

According to law of conservation of momentum, the change in momentum before collision will be equal to the change in momentum of the objects after collision in isolated system.

But as it is perfectly inelastic collision in the present case, the final momentum will be based on the product of total mass of both the object with the velocity with which the collision occurred. This form is attained from the law of conservation of momentum as shown below:

So as law of conservation of momentum,

                   M_{1} U_{1}+M_{2} U_{2}=M_{1} V_{1}+M_{2} V_{2}

Here M_{1} = 3 kg  and M_{2} = 2 kg are the masses of objects 1 and 2, U_{1} = 1.4 m/s  and U_{2} = 0 are the initial velocities of object 1 and object 2,  V_{1} and  V_{2} are the final velocities of the objects.

So after collision, object 1 get sticked to object 2 and move together with equal velocity V_{1} =  V_{2} = V_{f}. Thus the above equation will become,

            M_{1} U_{1}+M_{2} U_{2}=\left(M_{1}+M_{2}\right) V_{f}

So the final velocity is

              V_{f}=\frac{M_{1} U_{1}+M_{2} U_{2}}{\left(M_{1}+M_{2}\right)}

Thus,

       V_{f}=\frac{(3 \times 1.4+2 \times 0)}{(3+2)}=\frac{4.2}{5} = 0.84 ms-1.

8 0
4 years ago
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