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A lead bullet of mass 0.05kg is fired with a velocity of 200ms\(^{-1}\) into a lead block of mass 0.95kg. Given that the lead block can move freely, the final kinetic energy after impact is
50 J
100 J
150 J
200 J
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k.E=(mv^2)/2 . . .wia m=mass, v=linear velocity . .the values of d quantity mass z given which z sum of dia masses but v z nt given which z d common velocity of d two bodies. . .i proceed 2 find v wit d law of conservation of momentum which states dat initial momentum=final momentum. . mv +mv=V(m+m) . .subtituting, 0.05*200 + 0.95*0=v(0.05+0.95) . .10+0=v(1). . .v=10m/s. . Then K.E=(mv^2)/2 ,subtituting ,K.E=({0.05+0.95}*10*10)/2 ,K.E=100/2=50J
from the principle of conservation of linear momentum.. M1V1- M2V2 = (M1 + M2)V
where M1= 0.05, M2= 0.95, V1=200, V2=0, then V which is the common velocity was not given I.e. V=?...
therefore: 0.05×200 - 0.95×0= (0.05 + 0.95)V
=10 - 0 =(1)V
=V=10...
so... kinetic energy K.E= 1/2MV^2
NOTE: M= M1 + M2= 0.05 + 0.95=1
therefore:
K.E= 1/2 × 1 10^2
= 1/2 × 100
=50J.
I hope this is clear... thank u
M1V1 = M2V2
M1 = 0.05
V1 = 200
M2 = 0.95+ 0.05 = 1
V2 = ??(unknown)
Therefore V2 = M1*V1/M2
V2 = 0.05*200/1
V2 = 10
Kinetic energy = 1/2mv²
And kinetic energy of the final velocity is the kinetic energy of M2V2
= 1/2*1*10²
= 100/2
= 50
4rm d conservation of linear momentum which states total momentum b4 collision equal to total momentum afta collision,(M1U1) - (M2U2)=(M1+ M2)V,Given,M1-0.05,U1-200,M2-0.95,U2-0,V-?
the explanation are not clear and properly broken done
formula solving too are not also clear
From principle of conservation of linear momentum,
(0.05 x 200) + (0.95 x 0) = (0.05 + 0.95) x V (since collision is inelastic).
10 + 0 = V
Thus V = 10m/s.
Recall Kinetic Energy = \(\frac{1}{2} mv^2\)
\(\therefore\) K.E = 1/2 (0.05 + 0.95) x 10\(^2\)
K.E = 1/2 (1 x 100) = 50 J.
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1st of all,we wud find d velocity of both masses.n it's fig. 10,den Since it is d conservatn after d collitn,den it wud be 1/2(0.05 +0.95)*10square=50.
1st find d velocity ratio,usin d principle of linear momentum.After findin d velocity(v=m1v1 - m2v2 / m1 + m2) whc will gve 10m/s. Den will luk 4 d kinetic energy whc is 1/2mv^2 and ur final answer will gve u 50J.