Question

A block of mass m -2.00 kg collides head on with a block of mass m- 10.0 kg initially at rest on a rough horizontal surface. The block m strikes m2 with a speed of 4.00 m/s. Immediately after the very brief inelastic collision, the 2.00 kg block bounces back with a speed of 1.20 m/s. Ignore the effect of friction during the collision. (a) 4pts.] Calculate the speed of the 10.0 kg block immediately after the collision. (b) 4 pts.] Calculate the speed of the center of mass of the two blocks just before and also immediately after the colisol if tnetc fric (c) [4 pts.] If the collision lasts 4.00 ms, calculate the average force on each block during the collision. (d) Apts.) If the coefficient of kinetic friction between the blocks and the surface is p -0.250, alculate how far apart the two blocks will be when they stop. l4 pts.] Calculate the acceleration of the center of mass of the two blocks caused by the frictional forces after the collision (e)

Answer 1

a)

m₁ = 2 kg

v_1i = initial velocity before collision for m₁ = 4 m/s

v_1f = final velocity after collision for m₁ = - 1.20 m/s

m₂ = 10 kg

v_2i = initial velocity before collision for m₂ = 0 m/s

v_2f = final velocity after collision for m₂ = ?

Using conservation of momentum

m₁ v_1i + m₂ v_2i = m₁ v_1f + m₂ v_2f

(2) (4) + (10) (0) = (2) (- 1.20) + (10) v_2f

v_2f = 1.04 m/s

(b)

v_i = speed of center of mass before collision = (m₁ v_1i + m₂ v_2i )/(m₁ + m₂) = ((2) (4) + (10) (0))/(2 + 10) = 0.67 m/s

v_f = speed of center of mass after collision = (m₁ v_1f + m₂ v_2f )/(m₁ + m₂) = ((2) (- 1.20) + (10) (1.04))/(2 + 10) = 0.67 m/s

c)

For block m₁ :

F₁ = average force on m₁

t = time for collision = 4 ms = 0.004 sec

Using impulse-change in momentum equation

F₁ t = m₁ (v_1f - v_1i )

F₁ (0.004) = (2) (- 1.20 - 4)

F₁ = - 2600 N

For block m₂ :

F₂ = average force on m₂

t = time for collision = 4 ms = 0.004 sec

Using impulse-change in momentum equation

F₂ t = m₂ (v_2f - v_2i )

F₂ (0.004) = (10) (1.04 - 0)

F₂ = 2600 N

d)

Stopping distance due to frictional force is given as

d = v² /(2$\mu$g)

After collision, for block m₁ :

v_1f = speed of block m1 after collision = 1.20 m/s

d₁ = distance traveled by m₁ before stopping = ?

d₁ = v_1f² /(2$\mu$g) = (1.20)²/(2 (0.250) (9.8)) = 0.294 m

After collision, for block m₂ :

v_2f = speed of block m₂ after collision = 1.04 m/s

d₂ = distance traveled by m₁ before stopping = ?

d₂ = v_2f² /(2$\mu$g) = (1.04)²/(2 (0.250) (9.8)) = 0.221 m

Distance between the two blocks is given as

d = d₁ + d₂    (Since they travel in opposite direction after collision)

d = 0.294 + 0.221

d = 0.515 m