a 10.0-kg mass is travelling to the right with a speed of | studysoup
problem 73p a 10.0-kg mass is travelling to the right with a speed of 2.00 m/s on a smooth horizontal surface when it collides with and sticks to a second 10.0-kg mass that is initially at rest but is attached to a light spring with force constant 110.0 n/m. (a) find the frequency, amplitude, and period of the subsequent oscillations. (b) how long does it take the system to return the first time to the position it had immediately after the collision?
physics 101 notes: 8/25 (week 2 continued)  force o defined as a push or a pull, and is a vector quantity, meaning that it has direction depending on which way it is being pushed or pulled o it has amount and direction (how much force is being applied and in which direction it is moving)  newton’s first law o first draft: “an object that is free of forces move in a straight line and covers equal distances in equal time” o final: “an object that is free of external forces moves at a constant velocity”  how does a skater start, stop, or turn? o acceleration: “the rate at which an object’s velocity is changing with time o acceleration, s
u (initial velocity) = 18 km/hr 
t ( time )= 2.5 sec
v( final velocity) = 0 m/s
from, the first equation of motionv = u +  a×t0 = 5 +  a×2.5a = - 2 m/s2 hint: here, we will proceed by finding out the relative speed between the two cars and then using the formula i.e., time taken$ = \dfrac{{{\text{distance travelled}}}}{{{\text{speed}}}}$ to determine after how many hours these cars will meet each other.complete step-by-step answer:given, distance between the two cars d=160 kmstarting time for both the cars when they left their respective initial positions a and b is 8:10 amspeed of the first car, x=50 km per hourspeed of the second car, y=30 km per hourin order to obtain the relative speed between these two cars which are moving towards each other (i.e., in opposite directions), the individual speeds of these cars will be added.so, the relative speed between the two given cars, r= x+y=50+30=80 km per houras we know that, time taken$ = \dfrac{{{\text{distance travelled}}}}{{{\text{speed}}}}$so, time when both the cars will meet=$\dfrac{{{\text{initial distance between both the cars}}}}{{{\text{relative speed between the cars}}}} = \dfrac{{\text{d}}}{{\text{r}}} = \dfrac{{160}}{{80}} = 2$ hours.after 8:10 am, both the cars which are initially at different places a and b will meet each other in the next 2 hours.therefore, time when they will meet each = 8:10 am + 2 hours = 10:10 amhence, option a is correct.note: in this particular problem, since the given two cars are moving in opposite directions (or moving towards each other) that’s why the individual speed of the cars are added in order to determine the relative speed between them but if they were moving in the same direction then the relative speed between these cars would have been obtained by subtracting the individual speed of these cars. (a) f = 1/t = 2.11 hz

(b) 2*pi*f = 13.26 radians/s = sqrt(k/m)
m = 0.296 kg
solve for k
(c) (1/2)ka^2 = 0.150 j
solve for amplitude a, in meters (a) to my eye, they never have a common speed. that would require the two curves to have the same slope at the same time, which doesn't appear to me to be true.

(b) b is ahead of a but is slowing down and being passed by a, which has been moving at constant speed.

(c) b has an acceleration with greater magnitude. (in fact, the curve for a doesn't show any acceleration at all.)

(d) a has a constant slope, and therefore a constant velocity (and therefore displays no acceleration at all). b starts ahead of a initially with a velocity greater, but then decelerates as it nears t = 4 s. at around t = 3.7 s, b's velocity (slope) has nearly dropped to zero, and it gets passed by a.

if you find this helpful, please select favorite answer! a bob of mass m is suspended by inextensible string of length l from a fixed point. the bob is given a speed of at the bottom point. find the tension in the string when string deflects through an angle 120 degree from vertical. hi trinity, let's do this one!newton's famous second law states, in math terms: f = ma applying to the first case, we have: 100 = m(4.00) => m = 100/4 = 25 kgnow apply the formula to the second case: f = 25 x 16 = 400 newtons the value of head loss due to friction using chezy formula with constant c= 60 for a fluid, flowing through a 30 cm diameter and 70 m long pipe. velocity of fluid inside the pipe is 5 m/s. &nbsp;an electron enters uniform electric field maintained by parallel plates and of value 'e' &nbsp; v m-1 with a velocity ' v ' ms-1.&nbsp;the plates are separated by a distance 'd' metre. what is the acceleration of the electron in the field?&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;&nbsp; coulomb's law:f=kq1q2r2, where k is the coulomb's constant and r is the separation between the two charges. the magnitude of the net force:fnet=fx2+fy2k = 9 × 109 n·m2/c2the force due to charge q1 is:f1a=kq1qar1a2=(9×109)(21.4×10-6)(13.5×10-6)(80.1×10-2)2[readmore]f1a = 4.05 nthe force due to charge, q2:f2a=kq2qar2a2=(9×109)(30.6×10-6)(13.5×10-6)[(80.1×10-2)2+(80.1×10-2)2]2f2a = 2.9 nforce due to charge, q3:f3a=kq3qar3a2=(9×109)(57.3×10-6)(13.5×10-6)(80.1×10-2)2f3a = 10.9 nthe net force along the x-axis:fx=f1a-f2acos(45°)=4.05-(2.9)cos(45°)fx = 2.0 nthe net force in the y-direction:fy=f3a-f2acos(45°)=10.9-(2.9)cos(45°)fy = 8.85 nthe magnitude of the net force:fnet=fx2+fy2=2.02+8.852fnet = 9.07 nthe magnitude of the net force on the charge at point a is 9.07 n. 1.6 x 10-11