Marble A, mass 5.0 g, moves at a speed of 20.0 cm/s. It collides with a second marble, B, mass 10.0 g?

Question

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Marble A, mass 5.0 g, moves at a speed of 20.0 cm/s. It collides with a second marble, B, mass 10.0 g, moving at 10.0 cm/s in the same direction. After the collision, marble A continues with a speed of 8.0 cm/s in the same direction.

a. Calculate the marbles' momenta before the collision.
b. Calculate the momentum of marble A after the collision.
c. Calculate the momentum of marble B after the collision.
d. What is the speed of marble B after the collision.

Answer 1

change to m/s
20cm/s => .2m/s

Marble A
momentum = mass * velocity
momentum = (.005kg)(.2m/s)
momentum = .001 kgm/s

Marble B
momentum = (.01kg)(.1m/s)
momentum = .001 kgm/s

b) momentum = mass * velocity
momentum = (.005kg)(.08m/s)
momentum = 4 x 10^-4 kgm/s

c) find the velocity of Marble B after the collision
(.005kg)(.2m/s) + (.01kg)(.1m/s) = (.005)(.08m/s) + (.01kg)(v)
.002 = 4x10^-4 + .01v

.0016 = .01v
v= .16m/s

momentum = (.01kg)(.16m/s)
momentum = .0016kgm/s

d) in part c

Answer 2

You might want to draw a picture of what's going on. Basically, marble A and B move in the same direction. Marble A overtakes B, and transfers momentum to B.

Momentum = mv (this a vector quanity, so direction must be specified).
Prior to the collosion, for A : M = 5 x 20= 100 g-cm/sec
for B: M = 10 x 10 = 100 g-cm/sec
After the collision, for A: M = 5 x 8 = 40 g-cm/sec
Since momentum is conserved in a non-fricitional setting, for B: M = 160 g-cm/sec (100+100)=(40 + x)
and for B: v= 160 g-cm.sec /10 g = 16 cm/sec

Answer 3

You need to use the conservation of moment equation:
mA*vAi+mB*vBi = mA*vAf+mB*vBf

Since the marbles (probably) do not deform when they hit, this equation is valid.

a) just take the mass of each marble times the velocity of each marble

b) you know the final velocity of marble A and the mass so just multiply them together

c) the total momentum must remain the same on each side of the equation, so since you know the entire left hand side, and 1 of the terms on the right hand side, you just solve for the other term as a whole.

d) Take the answer from c) and divide it by the mass of B

Answer 4

m1 = 0.2kg m2 = 0.1kg v10 = 0.3 m/s v20 = 0.a million m/s v2f = 0.26 m/s v1f = ? Use Conservation of momemtum m1v10 + m2v20 = m1v1f + m2v2f, and make certain for v1f: v1f = (m1v10 + m2v20 - m2v2f)/m1 = 0.22 m/s it is diverse than the above solutions because i did not assume a completely inelastic collision. there grow to be no aspect out of both bodies sticking mutually, basically that they were transferring interior of an similar route in the previous and after the collision. through the way, the collision does no longer take care of mechanical power. you'll discover that about 128 mJoules of kinetic power is lost in the course of the collision, to different kinds of power inclusive of warm temperature, sound, etc.

Answer 5

Marble C