Answer:
Let M1 = 8 kg and M2 = 34 kg
F = M a = (M1 + M2) a
F = M2 g the net force accelerating the system
M2 g = (M1 + M2) a
a = M2 / (M1 + M2) g = 34 / (42) g = .81 g = 7.9 m/s^2
Answer:
(a)0.531m/s
(b)0.00169
Explanation:
We are given that
Mass of bullet, m=4.67 g=
1 kg =1000 g
Speed of bullet, v=357m/s
Mass of block 1,
Mass of block 2,
Velocity of block 1,
(a)
Let velocity of the second block after the bullet imbeds itself=v2
Using conservation of momentum
Initial momentum=Final momentum
Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s
(b)Initial kinetic energy before collision
Final kinetic energy after collision
Now, he ratio of the total kinetic energy after the collision to that before the collision
=
=0.00169
Answer:
<em>Answer: positive velocity & negative acceleration</em>
Explanation:
<u>Accelerated Motion</u>
Both the velocity and acceleration are vectors because they have magnitude and direction. When the motion is restricted to one dimension, i.e. left-right or up-down, the direction is marked with the sign according to some preset reference.
The locomotive is moving at a certain speed with a (so far) unknown sign but the acceleration has a negative sign. Since the locomotive comes to a complete stop it means the velocity and the acceleration are of opposite signs.
Thus the velocity is positive.
Answer: positive velocity & negative acceleration
Acceleration=(speed end - speed start)/ time
Data:
speed end=4 m/s
speed start=0 m/s
time=2.5 s
acceleration=(4 m/s - 0 m/s)/2.5 s=1.6 m/s²
Answer: the acceleration would be 1.6 m/s²
Answer:
Explanation:
A )
At the bottom of the circle , the potential energy of the stopper is converted into kinetic energy
1/2 m V² = mg x 2r + 1/2 mv²
m is mass of stopper , V is velocity at the bottom , r is radius of the circular path which is length of the string , v is velocity at the top
1/2 V² = g x 2r + 1/2 v²
V² = g x 4r + v²
V² = 9.8 x 4 + 8²
V² = 103.2
V = 10.16 m/s
B )
If T be the tension at the top
Net downward force
= mg + T . This force provides centripetal force for the circular motion
mg +T = mv² / r
T = mv²/r -mg
= m ( v²/r - g )
= .005 ( 8²/1 -g )
= .005 x 54.2
= .27 N .
C ) At the bottom
Net force = T - mg , T is tension at the bottom , V is velocity at bottom
T-mg = mV²/r
T = m ( V²/r +g )
= .005 ( 10.16²/1 +9.8)
= .005 x 113
= .56 N .
