Physics A 1.7 kg body is initially moving northward at 15 m/s.?

Question

A 1.7 kg body is initially moving northward at 15 m/s. Then a force of 20 N, toward the east, acts on it for a time of 4.5 s.

(a) At the end of that time, what is the body's final velocity?
Magnitude________m/s
Direction_________° north of east
(b) What is the change in momentum during that time?
__________kg*m/s

Answer 1

Answering (b) first,
∆p = (20 N)(4.5 s) = 90 kg•m/s
Then Ve = (90 kg•m/s)/(1.7 kg) ≈ 52.94118 m/s
θ ≈ arctan(15/52.94118) ≈ 15.81919 °N of E
and
V ≈ 15/sin(15.81919°) ≈ 55.02516 m/s

Since all given data have 2 significant digits,
θ ≈ 16°N of E
V ≈ 55 m/s

Answer 2

F = ma
a = F/m = 20/1.7 = 11.765 m/sec/sec
v along x = 0 + 4.5(11.765) = 52.94m/sec
v along y = 15 m/sec
lvl = (15^2 + 52.94^2)^(1/2) = 55.02 m/sec
Direction = arctan(15/52.9) = 15.83 deg north of east
Momentum change = 1.7( 55.02 - 15) = 68.034

Answer 3

F=ma, a=F/m

a = 20N/1.7kg = 11.76m/s^2 east

a = dv/dt = V2-V1/t
t = 4.5s
a = 11.76
V1 = 0

solve for V2 = 52.94 m/s east

magnitude = sqrt (15^2 + 52.94^2)
dir = arctan (15/52.94)
momentum change - multiply the magnitudes of velocity times the mass, ez to find.