Can someone explain why do the signs change on the final answer? I just need an explanation because I got the answer here. I`m confused because the answer was negative at first then it becomes positive.a 3.Using Momentum Consnvation,* mi4 tm2Mz:. 3x4) + 8x(-2) = (3+8) vV= -0-364m/s]→ 0:344m/3 towards left.- (m, tm2) VMomenkum consuvation,Usingmidl, tmqlle = m,v, tone Vz3x4 + 8x(m2) - 3(-1) + 8 (V):. V2 = -0.128 m/s)→ 6.125 m/s towands ut. Given dataMass of 1st ball (ml) = 200 gMass of 2nd ball(m2) = 400 gSpeed of 1st ball (u) = 30 cm/sSpeed of 2nd ball be vSpeed of both ball after collision is zero(motionless)SolutionConservation of momentumInitial momentum = final momentummlu + m2v =0200 x 30 + 400 x v=0v= - 15 cm/snegative shows that 2nd ball moves to leftwardv= 15 cm/s

Answer is velocity = -15 m/s Here , negative sign signifies the direction of velocity . And 15…

Can someone explain why do the signs change on the final answer? I just need an explanation because I got the answer here. I`m confused because the answer was negative at first then it becomes positive.a

Can someone explain why do the signs change on the final answer? I just need an explanation because I got the answer here. I`m confused because the answer was negative at first then it becomes positive.a

University Physics Volume 3

17th Edition

ISBN:9781938168185

Author:William Moebs, Jeff Sanny

Publisher:William Moebs, Jeff Sanny

Chapter9: Condensed Matter Physics

Section: Chapter Questions

Problem 51P: The Csl crystal structure is BCC. The equilibrium spacing is approximately r0=0.46nm . If Cs+ ion a...

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Proper Length

The theory of special relativity explains the connection between space and time to objects that move in a straight line at a constant speed. The objects move at the speed of light. When an object approaches light speed, its mass is endless and cannot move faster than light. The most crucial postulate of Einstein's special relativity is that the laws of nature do not change in each frame of reference.

Question

3.
Using Momentum Consnvation,
* mi4 tm2Mz
:. 3x4) + 8x(-2) = (3+8) v
V= -0-364m/s]
→ 0:344m/3 towards left.
- (m, tm2) V
Momenkum consuvation,
Using
midl, tmqlle = m,v, tone Vz
3x4 + 8x(m2) - 3(-1) + 8 (V)
:. V2 = -0.128 m/s)
→ 6.125 m/s towands ut.

Given data
Mass of 1st ball (ml) = 200 g
Mass of 2nd ball(m2) = 400 g
Speed of 1st ball (u) = 30 cm/s
Speed of 2nd ball be v
Speed of both ball after collision is zero(motionless)
Solution
Conservation of momentum
Initial momentum = final momentum
mlu + m2v =0
200 x 30 + 400 x v=0
v= - 15 cm/s
negative shows that 2nd ball moves to leftward
v= 15 cm/s

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