Impulse and Momentum A block with mass 4 kg slides 1.2m from rest down a ramp of
ID: 1433666 • Letter: I
Question
Impulse and Momentum
A block with mass 4 kg slides 1.2m from rest down a ramp of angle 25°. It then leaves the ramp and slides along a level frictionless surface until it collides with a second block of mass 7 kg that is at rest. After the collision, if the first block rebounds with a speed of 1.8 m/s backwards.
a) What was the impulse felt by the first block?
b) What was the impulse felt by the second block?
c) If the collision lasts 200ms, what is the force felt by each block during the collision?
d) With what velocity is the second block moving after the collision?
Explanation / Answer
Speed of first block as it leaves ramp,
Using Energy Conservation
Initial Potential Energy = Final Kinetic Energy
m*g*h = 1/2*mv^2
9.8*1.2*sin(25) = 1/2*v^2
v = 3.15 m/s
Now,
Using Momentum Conservation,
Initial Momentum = Final Momentum
m1*v1 + m2*v2 = m1*v1f + m2*v2f
4*3.15 + 0 = - 4*1.8 + 7*v2f
v2f = 2.83 m/s
(a)
Impulse felt by the first block, = m*V
Impulse felt by the first block, = 4 * (-1.8 - 3.15)
Impulse felt by the first block, = - 19.8 kgm/s
(-ve sign shows the direction toward left)
(b)
Impulse felt by the second block, = m*V
Impulse felt by the second block, = 7 * (2.83 - 0)
Impulse felt by the second block, = 19.8 kgm/s
(c)
Impulse, F*t = m*V
F * 200 * 10^-3 = 19.8
F = 99 N
Force felt by each block during the collision, F = 99 N
(d)
Velocity of the second block moving after the collision, v2f = 2.83 m/s