Question

Blocks A (mass 4.00 kg ) and B (mass 6.00 kg ) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 5.00 m/s . The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. Let x be the direction of the initial motion of block Find the maximum energy stored in the spring bumpers. Find the velocity of block A when the energy stored in the spring bumpers is maximum. Find the velocity of block B when the energy stored in the spring bumpers is maximum. Find the velocity of block A after they have moved apart. Find the velocity of B after they have moved apart.

Answer 1

Answer and Explanation:

mass of the bock A is, m1=4 kg

mass of the bock B is, m2=6 kg

velocity of the block A is u1=5 m/sec

velocity of the block B is u2=0

a)

by using law of conservation of energy,

m1*u1+m2*u2=(m1+m2)*v

4*5+0=(4+6)*v

final velocity v=2 m/sec ( when the spring gets compression)

by using law of conservatio of enegry,

maximum energy stored U=K1-K2

U=1/2*m1*u1^2-1/2*(m1+m2)v^2

U=1/2*4*5^2-1/2*(4+6)*2^2

U=30 J

b)

if energy stored in the bumper is maximum,

velocity of block A is, v=2 m/sec

velocity of block B is, v=2 m/sec

c)

after moving back,

velocity of the block A is,

v1=((m1-m2)/(m1+m2))*u1

v1=((4-6)/(4+6))*5

v1=-1 m/sec

and

velocity of the block B is,

v2=((2*m1)/(m1+m2))*u1

v2=((2*4)/(4+6))*5

v2=4 m/sec

Answer 2

The maximum energy stored in the bumper is 50 J.

The velocity of block A when  the energy stored in the spring bumpers is maximum is 5 m/s.

The velocity of block B when  the energy stored in the spring bumpers is maximum is 0.

The final velocity of block A after they have moved apart is -1 m/s.

The final velocity of block B after they have moved apart is 4 m/s.

The given parameters;

• mass of block A = 4 kg
• mass of block B = 6 kg
• initial velocity of block B = 0
• initial velocity of block A = 5 m/s

The maximum energy stored in the bumper is determined by applying the principle of conservation of energy;

$U_x = K.E_A + K.E_B\\\\U_x = \frac{1}{2} mv^2_A + \frac{1}{2} mv^2_B\\\\ U_x = \frac{1}{2} \times 4 \times 5^2 \ + \ \frac{1}{2} \times 6 \times 0\\\\U_x = 50 \ J$

The velocity of block B when  the energy stored in the spring bumpers is maximum = 0

The velocity of block A when  the energy stored in the spring bumpers is maximum = 5 m/s

The velocity of both blocks when they move apart is calculated by applying the principle of conservation of linear momentum;

$m_1u_1_a + m_2u_2 _b = m_1v_1a + m_2 v_2 _b\\\\4(5) \ + \ 6(0) = 4v_1_a + 6v_2_b\\\\20 = 4v_1_a + 6v_2_b \ ---(1)$

Apply one directional velocity formula;

$u_1_a + v_1_a = u_2_b + v_2 _b\\\\5 + v_1_a = 0 + v_2 _b\\\\5 + v_1_a = v_2_b \ ---(2)$

substitute in the value of $v_2_b$ in (1);

$20 = 4v_1_a +6(5 + v_1a)\\\\20 = 4v_1_a + 30 + 6v_1a\\\\-10= 10v_1_a\\\\-1 \ m/s = v_1_a$

The final velocity of block B is calculated as;

$v_2_b = 5 + (-1)\\\\v_2_b = 4 \ m/s$

Learn more here:https://brainly.com/question/24424291