Two ice skaters approach each other at right angles. Skater A has a mass of 32.7 kg and travels in the x direction at 2.3 m/s. S
1 answer:
Answer:
Final speed, v = 1.25 m/s
Explanation:
Given that,
Mass of skater A,
Initial speed of skater A, (x axis)
Mass of skater B,
Initial speed of skater B, (y axis)
It is mentioned that the two skaters collide and cling together. It is case of inelastic collision in which momentum remains conserved. Let V is the final speed of the couple. It is given by :
So, the final speed of the couple is 1.25 m/s. Hence, this is the required solution.
You might be interested in
Answer:
(a) 0.38 m
(b) 2.78 m/s
(c) 0.11 watt
Explanation:
mass, m = 0.3 kg
spring constant, K = 160 N/m
initial compression, d = 12 cm = 01.2 m
initial speed, u = 3 m/s
(a) Let the maximum stretch is y.
Use conservation of energy
Initial potential energy + initial kinetic energy = final potential energy
0.5 x K x d² + 0.5 x m x u² = 0.5 x K x y²
160 x 0.12 x 0.12 + 0.3 x 0.12 x 0.12 = 160 x y²
2.304 + 0.00432 = 160 y²
y = 0.38 m
y = 38 cm
(b) Let v is the maximum speed.
The speed is maximum when the stretch in the spring is zero, so by use of conservation of energy
Initial potential energy + initial kinetic energy = final kinetic energy
0.5 x K x d² + 0.5 x m x u² = 0.5 x m x v²
160 x 0.12 x 0.12 + 0.3 x 0.12 x 0.12 = 0.3 x v²
2.304 + 0.00432 = 0.3 v²
v = 2.78 m/s
(c) The time period of the spring mass system is given by
T = 0.272 second
Energy dissipated per cycle = 0.03 J
Power, P = 0.03 / 0.272 = 0.11 Watt