Answer:
v = 14.32 m/s
Explanation:
According to the principle of conservation of linear momentum, both the momentum and kinetic energy of the system are conserved. Since the two balls are in the same direction of motion before collision, then;
+
= (
+
) v
0.035 × 12 + 0.120 × 15 = (0.035 + 0.120) v
0.420 + 1.800 = (0.155) v
2.22 = 0.155 v
⇒ v = 
= 14.323
The velocity of the balls after collision is 14.32 m/s.
Answer:
the person will be in the shore at 10.73 minutes after launch the shoe.
Explanation:
For this we will use the law of the lineal momentum.

Also,
L = MV
where M is de mass and V the velocity.
replacing,

wher Mi y Vi are the initial mass and velocity, Mfp y Vfp are the final mass and velocity of the person and Mfz y Vfz are the final mass and velocity of the shoe.
so, we will take the direction where be launched the shoe as negative. then:
(70)(0) = (70-0.175)(
) + (0.175)(-3.2m/s)
solving for
,
= 
= 0.008m/s
for know when the person will be in the shore we will use the rule of three as:
1 second -------------- 0.008m
t seconds-------------- 5.15m
solving for t,
t = 5.15m/0.008m
t = 643.75 seconds = 10.73 minutes
Answer:
t = 0.657 s
Explanation:
given,
initial vertical velocity = 7.5 m/s
initial horizontal velocity = 0 m/s
angle = 49◦
using kinetic equation
final velocity in vertical direction
v sinθ = u_y - gt ........................(1)
final velocity in horizontal direction
v cosθ = u_x + a_x × t
here u_x = 0.0 m/s
v cosθ = a_x×t ......................(2)
Dividing equation (1) / (2)

solving for time t

u_y = initial velocity along x direction
acceleration along a_x = 1.4 m/s²
g = acceleration due to gravity = 9.8 m/s²
θ = 43° , u_y = 7.5 m/s

t = 0.657 s
time taken by the particle is t = 0.657 s