Answer:
1.196kgm/s
Explanation:
The change in momentum [tex]\delta\rho[/tex] of the arrow is given by;
[tex]\delta\rho=mv-mu...............(1)[/tex]
where m is the mass of the arrow, u is the initial velocity and v is the final velocity.
Given;
m = 26.0g = 0.026kg
u = 0 since the arrow was initially at rest before being released.
v = 46m/s.
Since the intial velocity is zero, equation (1) reduces to
[tex]\delta\rho=mv...............(2)[/tex]
Hence the change in momentum is calculated thus;
[tex]\delta\rho=0.026kg*46m/s\\\delta\rho=1.196kgm/s[/tex]
