Two ice skaters push off against one another starting from a stationary position. The 45.0 kg skater acquires a speed of 0.375 m/s. What speed does the 60.0 kg skater acquire? Assume that any unbalanced forces during the collision are negligible. 0.500 m/s 0.281 m/s 0.375 m/s 0.750 m/s 0.000 m/s An 8.0-g bullet is shot into a 4.0-kg block, at rest on a frictionless horizontal surface (see the figure). The bullet remains lodged in the block. The block moves into an ideal massless spring and compress it by 3.7 cm. The spring constant of the spring is 2500 N/m. The initial velocity of the bullet is closest to 460 m/s 440 m/s 48 m/s 500 m/s 520 m/s (Draw a free-body diagram for the block) A 50.0-kg block is being pulled up a 16.0 degree slope by a force of 250 N which is parallel to the slope. The coefficient of kinetic friction between the block and the slope is 0.200. What is the magnitude of the acceleration of the block? 0.528 m/s^2 0.158 m/s^2 0.412 m/s^2 0.983 m/s^2 0.260 m/s^2 In the figure, a mass of 31.77 kg is attached to a light is wrapped around a cylindrical spool of radius 10.0 cm and moment of inertia 4.00 kg m^2. The spool is suspended from the ceiling, and the mass isthen released from rest a distance 5.70 m above the floor. How long does it take to reach th floor? 7.89 s 3.83 s 3.98 s 5.59 s 1.14 s
Masses m = 45 kg
M = 60 kg
Speed of 45 kg skater v = 0.375 m/s
From law of conservation of linear momentum,
mv +MV = 0
MV = -mv
V = -mv /M
= -(45x0.375)/60
= -0.28125 m/s
So, answer is 0.28125 m/s
(25).mass of bullet m = 8 g = 0.008 kg
mass of block M = 4 kg
Spring constant k = 2500 N/m
Compression X = 3.7 cm =0.037 m
Potential energy of the spring U =(1/2) kX 2
= 0.5x2500x0.037 2
= 1.71125 J
Kinetic energy of the bullet block system K = U
(1/2)(m+M) v 2 = 1.71125
v 2 = (2 x1.71125) /(m+M)
= 3.4225 /4.008
= 0.8539
v = 0.924 m/s
From law of conservation of momentum , mu +MU =(m+M) v
(0.008u ) + (4 x0) =(0.008+4)0.924
u = 462.96 m/s
= 460 m/s
Homework Sourse
