Answer:
(a)0.531m/s
(b)0.00169
Explanation:
We are given that
Mass of bullet, m=4.67 g=
1 kg =1000 g
Speed of bullet, v=357m/s
Mass of block 1,
Mass of block 2,
Velocity of block 1,
(a)
Let velocity of the second block after the bullet imbeds itself=v2
Using conservation of momentum
Initial momentum=Final momentum







Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s
(b)Initial kinetic energy before collision



Final kinetic energy after collision



Now, he ratio of the total kinetic energy after the collision to that before the collision
=
=0.00169
Answer:
10kg
Explanation:
Let PE=potential energy
PE=196J
g(gravitational force)=9.8m/s^2
h(change in height)=2m
m=?
PE=m*g*(change in h)
196=m*9.8*2
m=10kg
Answer:
a) 141.6m
b) 8.4m/s
Explanation:
a) to find the total displacement you use the following formula for each trajectory. Next you sum the results:

hence, the total distance is 141.6m
b) the mean velocity of the total trajectory is given by:

hence, the mean velocity is 8.4 m/s
Explanation:e=mwS
<em>and also yes it is i think </em>