A small block of mass 20.0 grams is moving to the right on a horizontal friction
ID: 1425164 • Letter: A
Question
A small block of mass 20.0 grams is moving to the right on a horizontal frictionless surface with a speed of 0.580 m/s. The block has a head-on elastic collision with a 40.0 gram block that is initially at rest. Since the collision is head-on, all velocities lie along the same line, both before and after the collision.
Two blocks are sliding on a horizontal frictionless surface with velocities shown in the sketch. Block A has mass 0.500 kg and block B has mass 0.350 kg. The two blocks have a perfectly inelastic collision and stick together after the collision. 0.200 m/s B 0.400 m/s (a) What is the speed of the combined blocks after the collision? m/s (b) What angle does the velocity of the combined blocks make with the +x-axis after the collision? | 0, counterclockwise from the +x-axis counterclockwise from the +x-axis (c) What is the magnitude of the decrease in kinetic energy of the system of two blocks due to the collision?Explanation / Answer
here,
mass of block A , mA = 0.5 kg and uA = 0.2 m/s i
mass of block B , mB = 0.35 kg and uB = 0.4 m/s j
(a)
let the speed of combined mass be v
using conservation of energy
mA * uA + mB * uB = ( mA + mB) * v
0.5 * 0.2 i + 0.35 * 0.4 j = 0.85 * v
v = 0.12 i m/s + 0.16 j m/s
|v| = sqrt( 0.12^2 + 0.16^2)
|v| = 0.2 m/s
the speed of combined blocks is 0.2 m/s
(b)
theta = arctan(0.16/0.12)
theta = 53.13 degree
the angle is 53.13 degree from the +x axis counterclockwise
(c)
the decrease in kinetic energy , K = 0.5 * mA * uA^2 + 0.5 * mB * uB^2 - 0.5 * (mA + mB) * v^2
K = 0.5 * ( 0.5 * 0.2^2 + 0.35 * 0.4^2 - 0.85 * 0.2^2)
K = 0.021 J
the decrease in kinetic energy is 0.021 J