a projectile of mass 50 gm is fired with velocity 10m s at an angle 30 with the horizontal calculate the linear momentumand kinetic energy of the projectile at highest point?
At the highest point the vertical component of the velocity will be zero, do you agree with this?
becoz it was fired horizantly
|dw:1341756910842:dw| If it was given only vertical velocity, then also at the highest point it'd be zero
at the highest point linear momentum would be mucos theta
\[v ^{2}= u ^{2} - 2as\] u is 10 sin theta, a is acc due to gravity, find s, then we use it to find potential energy which is mgs which we get 125 so energy is 125
can we do lik this? k.e= 1/2 m sq(u cos theta)
sorry it is 6.25, @mathavraj i thought it is safe to just see change in height
@mathavraj i think we can. ucos theta is the velocity at the highest point. sub it in formula for KE should give us the answer.
i got KE as 1.87 J
@Ruchi what is the answer?
@Vaidehi09 what did u do?
KE = 1/2 m u^2 cos^2(30)
it should be sin thta na?
theta is the angle with the horizontal. so horizontal component of velocity will have cos..not sin
no u sin theta is zero at top
yes and ke energy is chaange in velocity so it will be 1/2m(vi-vf)
pls explain @rdx
well kinetic energy is change in its intial energy and final energy so taking sin theta or cos theta should not make a difernece ,
so let me do this by taking cos component in consideration \[1/2m(v \cos 30-v \cos 0)^{2}\]
check it wid sin shoot me if u get any other result :P
@mathavraj any problem??
its zero wen we use sin component..but i still dint get that change thing..at the highst point there is only cos component...wich is constant throught the motion
umm wait
sorry i made a mistake again but 6.25 is correct. u urself agree that that there is no change in cos componet so there is no change in kinetic energy in x direction but only change in kinetic energy y direction and initially K.E in y direction is 6.25 and finally it is 0 so initial - fianl is the net K.E
so should we take sin also
what do u wnat to say?
@Ruchi. what is the answer?
i don't know its answer is nt mention
we don;t hav to use its mass
?
y not mass?
1/2"M"v^2
@Rdx ?
yes what do u want? u will need mass to find the energy, whenever a qustion asks potential or kinetic energy u actually need to calculate change in potential or kinetic energy
best is solve by seeing change in height
@Rdx KE at the highest point = KE in x direction + KE in y direction.....right? but velocity in y direc at that pt = 0 so KE at highest pt = KE in x direc........or am i wrong?
IT IS CHANGE IN KINETIC ENERGY,
well the x componet of velocity remains constant throughot and the change is y componet |dw:1341827827130:dw|
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