Question

Block A of mass, mA = 1.7 kg is shot from a spring device of spring constant, k = 700 N/m along a frictionless horizontal surface. The initial compression of the spring is 0.300 m. The shot makes the block rise to another horizontal level at a height h= 1m above the first. On this horizontal it collides with another stationary block B of mass mB = 3.5 kg. The blocks stick together and encounter a rough surface. The blocks eventually come to a stop after traveling a distance d = 1.85 m.

1. What is the velocity of block A before it collides with the block B?

2. What is the velocity of the two blocks right after collision?

3. Is the total kinetic energy of the system conserved in the collision of the blocks A and B? If not, how much kinetic energy is lost during the collision?

4. What is the impulse received by A during the collision?

5. What is the coefficient of kinetic friction on the rough surface?

6. What is the center of mass velocity of two blocks just before the collision?

Answer 1

intial elongation = 0.3 m and it rises to a height. therefore by energy conservation

POTENTIAL ENERGY = KINETIC ENERGY

Kr? + magH m4u 2 2

x 700 x 0.32 + 1.7 X 10 X 1= = x 1.7 x 12

on solving the above equation velocity of mass A just before hitting mass B

n v = 7.55- sec

(b) by momentum conservation

intial momentum = final momentum

MA X VAi + mg XuBi = (ma+mb) X Us

1.7 x 7.55 +3.5 x 0 = (1.7 +3.5) XUF

$v_{f}= 2.46 \frac{m}{sec}$

(c) kinetic energy is not conserved as the collision is inelastic

lost kinetic energy

$\Delta K.E= K.E_{f}-K.E_{i}$

$K.E_{f}=\frac{1}{2}(m_A+m_B)v_f^2$

$K.E_{f}=\frac{1}{2}(1.7+3.5)(2.46)^2$

K.E = 15.85 joule

$K.E_{i}=\frac{1}{2}m_A v_{Ai}^2+\frac{1}{2}m_Bv_{Bi}^2$

$K.E_{i}=\frac{1}{2}\times 1.7\times {7.55}^2+\frac{1}{2}\times 3.5\times 0^2$

$\Delta K.E=48.45- 15.85 =32.6 joule$

(4) impulse = change in momentum

$I= m_a \times v_f-m_a \times v_i$

$I= 1.7 \times ( 2.46-7.55)$

$I= 8.653 \frac{kg \times m}{sec}$

(5) at friction surface blocks stop after a distance of 1.85 m

by newtons law

$\mu mg =ma$

$\mu =\frac{a}{g}$

where a = accelaration

from newtons third equation

$v^2=u^2+2as$

$0^2=2.46^2+2a \times 1.85$

$a= -1.63\frac{m}{sec^2}$

$\mu = \frac{a}{g}=\frac{1.63}{10}$

$\mu = 0.163$

(6) velcity of centre of mass

$v_{com}=\frac{m_A \times v_{Ai}+m_B \times v_{Bi}}{m_A+m_B}$

$v_{com}=\frac{1.7 \times 7.55+3.5 \times 0}{1.7+3.5}$

$v_{com}=2.46 \frac{m}{sec}$