Answer:
Explanation:
My speed after the interaction will depend upon the impulse the ball will make on me . Now impulse can be expressed as follows
Impulse = change in momentum
change in momentum in the ball will be maximum when the ball bounces back with the same velocity which can be shown as follows
change in momentum = mv - ( - mv ) = 2mv
So when ball is bounced back with same velocity , it suffers greatest impulse from my hand . In return , it reacts with the same impulse on my hand pushing me with greatest impulse according to third law of motion. this maximizes my speed after the interaction.
Answer:
a) a = - 0.0833 m / s², b) t = 4.4 s
Explanation:
a) this is a kinematics exercise where the acceleration is along the inclined plane
v = v₀ - a t
a = v₀ - v / t
a = 3 - 8/60
a = - 0.0833 m / s²
b) in this case the final velocity is zero
v = v₀ - a t
0 = v₀ - at
t = v₀ / a
t = 28 / 6.4
t = 4.375 s
t = 4.4 s
Answer:
Explanation:
We shall apply conservation of momentum law in vector form to solve the problem .
Initial momentum = 0
momentum of 12 g piece
= .012 x 37 i since it moves along x axis .
= .444 i
momentum of 22 g
= .022 x 34 j
= .748 j
Let momentum of third piece = p
total momentum
= p + .444 i + .748 j
so
applying conservation law of momentum
p + .444 i + .748 j = 0
p = - .444 i - .748 j
magnitude of p
= √ ( .444² + .748² )
= .87 kg m /s
mass of third piece = 58 - ( 12 + 22 )
= 24 g = .024 kg
if v be its velocity
.024 v = .87
v = 36.25 m / s .
Wavelength = speed / frequency = 340 / 17000 = 0.02 m
