Asked by kate
A bullet with mass 4.91 g is fired horizontally into a 1.981-kg block attached to a horizontal spring. The spring has a constant 6.25 102 N/m and reaches a maximum compression of 6.38 cm.
a) find the initial speed of the bullet-block system.
b) find the speed of the bullet
a) find the initial speed of the bullet-block system.
b) find the speed of the bullet
Answers
Answered by
drwls
a) By "initial speed", they really mean "right after the bullet hits the block and gets embedded". Call that speed v. The bullet-block speed is zero before that.
Assume energy is conserved during spring compression.
(1/2)(M+m) v^2 = (1/2)k*X^2
v = sqrt [k/(M+m)]*X
= sqrt[625/1.985] s^-1 *0.0638 m
= 1.132 m/s
b) For the initial bullet speed V, apply conservation of momentum to the bullet-stopping process. Mechanical energy is NOT conserved then.
m*V = (m+M)*v
(0.00491 kg)* V = (1.985 kg)* v
V = 458 m/s
Assume energy is conserved during spring compression.
(1/2)(M+m) v^2 = (1/2)k*X^2
v = sqrt [k/(M+m)]*X
= sqrt[625/1.985] s^-1 *0.0638 m
= 1.132 m/s
b) For the initial bullet speed V, apply conservation of momentum to the bullet-stopping process. Mechanical energy is NOT conserved then.
m*V = (m+M)*v
(0.00491 kg)* V = (1.985 kg)* v
V = 458 m/s