# Soultions to Various Problems in Newtonian Mechanics

See attachment for problem please.

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4. Since the system is motionless before the string is burned, by conservation of momentum the total momentum of the blocks must remain zero after the string is burned, i.e. we have

m1 v1 + m2 v2 = 0.

Thus we have

v1 = -m2 v2/m1 = -(7.6)(4.6)/(0.28) = -124.9 m/s,

the minus sign meaning that v1 points to the left as shown. Now by conservation of energy, the potential energy E stored in the spring before the string is burned must be equal to the total kinetic energy of the blocks after the string is burned, i.e.

E = 1/2 (m1 v1^2 + m2 v2^2)

= 1/2[(0.28)(124.9)^2 + (7.6)(4.6)^2]

= 2260 J

All units are MKS, so there is no need to write them down in the calculation.

2. Note: The answer to this problem is not well-defined because it depends on where on the incline the portion with friction is located. We will assume that it is located at the top of the incline. Otherwise, the answer will be greater since the block will enter the portion with friction later and thus have a greater speed when decelerated by friction, and thus this constant deceleration will occur over a shorter period of time, slowing down the block less.

Let v1 be the speed of the block immediately after it leaves the spring and let v2 be its speed immediately after it leaves the portion of the incline with friction. By conservation of energy, we have

1/2 m v1^2 = 1/2 k x^2,

where x = 0.35 cm is the initial compression of the spring. Thus we have

v1 = x sqrt(k/m) = (0.35)sqrt(310/2) = 4.36 m/s.

Now as the block slides over the portion of the incline with friction, we have

ma = mg sin theta - mu N

= mg sin theta - mu mg cos theta

where a is the acceleration of the block (deceleration if negative) and theta = 15 degrees is the incline of the plane. Thus we have

a = g(sin theta - mu cos theta).

Plugging in the numbers, we find

a = (9.8)(sin 15 - 0.5 cos 15) = -2.20 m/s^2.

Now the speed of the block when it is sliding over the portion of the incline with friction is given by

v(t) = v1 - |a|t

where t is the time since the block started sliding down the incline. Thus we have

s(t) = v1 t - 1/2|a|t^2.

Letting T be the total time for the block to slide down the portion of the incline with friction, we have

d = v1 T - 1/2|a|T^2

where d = 0.9 m is the length of the portion of the incline with friction. Solving for T, we have

T = (v1 - sqrt(v1^2 - 2|a|d))/|a|.

Plugging in the numbers, we have

T = (4.36 - sqrt(4.36^2 - 2(2.20)(0.9))/(2.20) = 0.218 s.

Thus we have

v2 = v1 - |a|T

= 4.36 - (2.20)(0.218) = 3.88 m/s.

To find vf, we use conservation of energy once again. We have

(1) 1/2 m v2^2 + mg h2 = 1/2 m vf^2,

where h2 is the height of the block immediately after it leaves the portion of the incline with friction. We have

h2 = h1 - d sin theta,

where h1 = 50 cm is the height of the block at the top of the incline. Thus we have

h2 = 0.5 - 0.9 sin 15 = 0.267 m.

Solving (1) for vf, we find

vf = sqrt(v2^2 + 2 g h2)

= sqrt(3.88^2 + 2(9.8)(0.267))

= 4.50 m/s.

10. (part 1) Let y0 = 1097 m be the initial altitude of the skydiver and y1 = 193 m be his altitude after he opens up his chute. Before the chute is open, we have

m a0 = F0 - mg,

where a0 is the acceleration of the diver before he opens his chute, F0 = 50 N is the retarding force on the diver before his chute is open and m = 86 kg is his mass. Thus we have

a0 = F0/m - g = 50/86 - 9.8 = -9.22 m/s^2.

(Note that this acceleration is negative because the skydiver is accelerating toward the ground.) After the chute is open, we have

m a1 = F1 - mg

where a1 is the acceleration of the diver after he opens his chute and F1 = 1611 N is the retarding force on the diver after his chute is open. Thus we have

a1 = F1/m - g = 1611/86 - 9.8 = +8.93 m/s^2.

(Note that this acceleration is positive because the skydiver is now decelerating toward the ground.)

Now the velocity of the diver immediately before he opens his chute is given by

v0 = a0 T0,

where T0 is the time the diver falls before opening the chute. We also have

y1 - y0 = 1/2 a0 T0^2,

whence

T0 = sqrt(2(y1 - y0)/a0)

= sqrt(2(193 - 1097)/(-9.22))

= 9.90 s.

The velocity v1 of the diver immediately before he opens his chute is given by

v1 = a0 T0 = (-9.22)(9.90) = -91.3 m/s.

While the parachute is open, we have

v(t) = v1 + a1 t,

where t is the time since the chute was first open. Thus we have

y(t) = y1 + v1 t + 1/2 a1 t^2.

whence

0 = y(T1) = y1 + v1 T1 - 1/2 a1 T1^2

where T1 is the time the diver falls with the parachute open. Solving for T1, we find

T1 = (-v1 - sqrt(v1^2 - 2 a1 y1))/(2 a1)

= (91.3 - sqrt(91.3^2 - 2(8.93)(193)) / (2(8.93))

= 1.20 s.

Now the final velocity of the diver when he hits the ground is given by

v2 = v1 + a1 T1.

Plugging in the numbers, we have

v2 = -91.3 + (8.93)(1.20) = -80.6 m/s.

Thus, the skydiver will hit the ground with a fatal speed of 80.6 meters per second! (He should've opened up his chute earlier to avoid this catastrophe!)

(part 2) In order to solve for the altitude y1 at which the diver must open his chute in order to hit the ground at a safe 4 m/s, we must solve the following equation for y1:

v2 = 4 m/s = v1 - a1 T1

= v1 - a1(v1 + sqrt(v1^2 - 2 a1 y1))/(2 y1)

= sqrt(2 a0(y1 - y0)) - a1[(sqrt(2 a0(y1 - y0)) + sqrt(2 a0(y1 - y0) - 2 a1 y1))/(2 y1)].

Unfortunately I ran out of time on this problem.

13. The acceleration of the satellite due to the rocket is given by

a = F/m.

Thus, the change is speed is given by

delta_v = aT = FT/m,

where T is the time with which the rocket is being fired. Thus we have

T = m delta_v / F

= (83000)(0.55)/56

= 815 s (= 13 m 35 s).

14. Let mc = 52 kg denote the mass of the canoe, mw = 58 kg be the mass of the woman, vc = 4.5 m/s be her velocity after she jumps off, and vc be the recoil velocity of the canoe after she jumps. Then we have

pi = pf

0 = mc vc + mw vw

whence

vc = -mw vw/mc

= -(58)(4.5)/(52)

= -5.02 m/s (to the left).

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