1 A bullet of mass 14.0g is fired horizontally into a block mass 30kg, in which it remains embedded. The block is suspended by a fine string so that it forms a simple pendulum of period 3.30s, and is initially at rest. After impact, the pendulum is deflected through an angle of 11.5deg. FInd the speed of the bullet at impact. (Hint take T = 2 pi (l/g)^0.5 for simple pendulum)
2 When a driver of mass 80kg steps into a car of mass 920kg, the vertica; height of the car aboe the raod decreases by 2.0cm. If the car is driven over a series of equally spaced bumps, the amplitude of vibration becomes much larger at on particular speed.
Calculate the separation of the bumps if it occurs at a speed of 15ms-1.
Thanks to all in adv!
Hi,
Let
m: mass of bullet,
v: speed of bullet at impact,
M: mass of block,
l: length of string,
theta: angle of deflection,
T: period of pendulum.
The steps are:
1. Given T, we find l.
2. Assuming no heat is lost, kinetic energy of bullet is used to bring the block (together with the embedded bullet) up, so
(1/2)(mv^2) = (M + m)(g)(h), where
h = l - {l cos (theta)},
from which v can be found.
Thanks.
Cheers,
Wen Shih
Hi,
1. Consider F = kx,
where F is the force exerted by the driver, k is the spring constant, and x is the compression of the spring.
Find k.
2. Consider T = (2 pi) sqrt(m / k),
where T is the period of oscillation, k is the spring constant, and m is the mass of the driver.
Find T.
3. Distance of bump = T x v,
where v is the velocity of the car.
Thanks.
Cheers,
Wen Shih
Hi,
There is a good collection of problems for further practice at this site:
http://physics.ucsd.edu/students/courses/summer2008/managed/physics2a/documents/chap15.pdf
Thanks.
Cheers,
Wen Shih