Answer:
a) v, v
b) 2mv^2
c) Elastic collion
Explanation:
(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v). From momentum conservation in x-direction
Here x, y represent direction.They are not variable. 1 and 2 represent before and after.
2vm=v1xm+v2xm, we find v1x=v.
From momentum conservation in y-direction
0 =v1ym+v2ym, we findv1y=v.
(b) By energy conservation principle
Before: K=1/2m(2v)^2=2mv^2.
After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2
(c) The collision is elastic
Answer:
when the mass of the bottle is 0.125 kg, the average height of the beanbag is 0.35 m.
when the mass of the bottle is 0.250 kg, the average maximum height of the beanbag is 0.91m.
when the mass of the bottle is 0.375 kg, the average maximum height of the beanbag is 1.26m.
when the mass of the bottle is 0.500 kg, the average maximum height of the beanbag is 1.57m.
Explanation:
It’s C because just trust
Answer:
a = - 1.987 × 10⁶ ft/s²
t = 6.84 × 10⁻⁴ s
Explanation:
v₀ = 910 ft/s
x = 5 in.
relation v = v₀ - k x
v = 0 as body comes to rest
0 = 900 - 5k/12
k = 2184 s⁻¹
acceleration
where
(A) a = -k × v
at v= 910 ft/s
a = - 1.987 × 10⁶ ft/s²
(B) at x = 3.9 in.
v = 910 - 3.9(2184)/12
v = 200.2 m/s
t = 6.84 × 10⁻⁴ s
